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- CHAPTER 07
AlternatingCurrent
Chapter Analysis w.r.t, Lost 3 Yeor's Boord ExomsThe analysis given here gives you an analytical picture of this chapter and will helpyou toidentify the concepts of the chapter that are to befocussed more from exam point of view.
Number of Questions asked in lost 3 years
2015 2016 2017Delhi All India I Delhi All India Delhi All India
Very Short Answer (1 mark) 1 Q 1 Q lQ lQ-Short Type I Answer (2 marks)
Short Type II Answer (3 marks) lQ lQ I lQ lQ
Long Answer (5 marks) 1 Q lQ lQ
Value Based Questions (4 marks)
• In 2015, one question of 1marks based on Capacitor Reactance and one numericalquestion of 3 marks based on AC Circuit in both sets.
• In 2016, in Delhi set, one question of 1mark based on Quality Factor and one question of5 marks based on AC Source were asked. In All India set, only one question of 5 marksbased on AC Circuit were asked.
• In 2017, in Delhi set, numerical question of 3 marks based on L-C-R Circuit and onequestion of 5 marks based on and Step up Transformer were asked. In All India set, onlyone question of 5 marks based on AC Source was asked.
On the basis of above analysis, it can be said that from exam point of view AC Source, ACCircuit, L-C-R Circuit and Step up Transformer are most important concepts of the chapter.
[TOPIC 1] Introduction to Alternating Current1.1 Alternating Current (AC)An alternating current is the current whosemagnitude changes continuously with timebetween zero and a maximum value and whosedirection reverses periodically.
I I . . 2 I· 21t= osmrot => I=Iosm revt= osm-tT
where, ro = angular frequency in rad/s10 = peak value or maximum value of AC.
Average or Mean Value of ACIt is defined as the value of AC which would sendsame amount of charge through a circuit inhalf-cycle that is sent by steady current in the
. (. ). 210same tune. i.e, T /2 I av = - = 0.637101t
The 63.7% of peak value of AC gives average ormean value of AC.Mean value of AC(I m) is 63% of peak value ofAC(Io) over any half-cycle. In a complete cycle ofAC, the mean value of AC will be zero.
Effective Value or rms Value of ACIt is defined as the value of AC over a completecycle which would generate same amount of heatin a given resistors that is generated by steadycurrent in the same resistor and in the same timeduring a complete cycle.
IInns = Ji. = 0.70710 = 70.7% of 10
The 70.7% of peak value of current gives effectiveor rms value of AC.
1.2 Alternating emf orVoltage
It is the emf or voltage which varies in bothmagnitude as well as direction alternatively andperiodically. The instantaneous alternating emf isgiven by
v = Vosin rotV.
Vrms= Ji = 0.707Vo
or v = Vocosrot
or Vrms = 70.7% of Vo
2VoVav = - = 0.637Vo or
1tVav = 63.7% of Vo
Both AC voltage and AC current are representedby diagrams as shown below:
v = Vo cos cot
Inductive Reactance (XL)The effective resistance or opposition offered bythe inductor to the flow of current is calledinductive reactance and is denoted by XL.
(-Frequency of AC main
XL = roL = 21tflAlso for a given inductor, XL = (21tL) f=> XLocf
21tL = constantwhere, L = self-inductance.
206
Capacitive Reactance (Xc)The opposing nature of capacitor to the flowof alternating current is called capacitivereactance.
(-Frequency of AC main
1 1Xc=-=--
roC 21tfC
For a given capacitor,
Xc = (2~C}]~ Xc OC]1-- = constant
21tCwhere, C = capacitance for AC.
1.3 AC CircuitsIn an AC circuit, both emf and current changecontinuouslyw.r.t. time, so in circuit, we haveto calculate average power in complete cycle(0 ~ T). Pay = Vrma Irma cos $where, cos $ = power factor.Average power consumption in pure inductiveand pure capacitive circuit is equal to zerobecause
Phase difference, $ = ~2
~ Power factor = cos ~ = 02
Wattless CurrentThe current in an AC circuit when averagepower consumption in AC circuit is zero, isreferred as wattless current or idle current.In other words, in an AC circuit R = 0~ cos $ = o. i.e. in resistance less circuit, thepower consumed is zero.
Chopterwise CSSE Solved Papers PHYSICS
Phasor DiagromThe representation of AC current and voltage (of samefrequency) by rotating vectors is called phasor and thediagram representing these phasors is known asphasor diagram.The length of the vector represents the maximum orpeak value. The projection of the vectors on fixed axisgives the instantaneous value of alternating currentand voltage.
Types of AC CircuitsAC Through ResistorSuppose a resistor of resistance R is connected to anAC source of emf with instantaneous value (E) is givenby
E = EosinrotR10
""EAn AC voltage applied to a resistor
t rVV/l
V-/graph
E - - - - - - - - - - - - - - -,Eo,,,,,,CJlt '
t-Phasor diagram
Then,(i) Voltage and current are in same phase.
(ii) Maximum current, 10 = VoR
(...) I VrmsIII = --rms R(iv) If V=Vosinrot, then I = 10 sin rot.
CHAPTER 7 : Alternating Current
AC Through InductorSuppose an inductor with self-inductanceL is connected to AC source withinstantaneous emf E is given byE = Eo sinox
L[:JAn AC source connected
to an inductor
V-I graph
1E or 1 Eo
t-
10Phasor diagram
Then,(i) Inductive reactance, XL = roL = 21t/L
( ") I Vo VoII 0=-=-XL roL
(...) I - Vrms _ Vrms _ VoIII - --------
rms XL roL roLIi
(iv) Voltage leads the current by phase ~.2
(v) If V = Vo sin rot,
then I = Io sin (rot - ~ )
(vi) Power factor, cos <l> = cos ~ = 02
(vii) Average power consumption,Pay = Vrms Irms cos <l> = 0
207
AC Through CapacitorSuppose a capacitor with capacitance C is connected toan AC source with emf having instantaneous value E isgiven by E = Eo sinror
C
i~~Ler-JAn AC source connected to a capacitor
1VII
1
V-I graph
E·------------EoIQ _
t+-«Phasor diagram
Then,
(.) C . . X 1 l.I apacinve reactance, c = - = -- l.e.roC 21t/C
1Xc oc-
/(ii) Capacitor offers infinite reactance in DC circuit as
/=0... Xc=oo
( ... ) Vo Vo CIII Io = - = -- = Vorox; (l/roC)
(iv) Voltage lags behind the current by phase~.2
(v) If V = Vo sin rot, then I = Io sin (rot + ~)-
I = !.sL = Eo = Erms = Erms
rms Ii l/roe-Ii l/roC Xc
(vi) Power factor is minimum and equal to zero.(vii) Average power consumption (during a complete
cycle), Pay = Vrms Irms cos <l> = 0
208
An L - R Series AC Circuit
AC source'-----~'"'-r=:....::..::="----'y
.................... yt 1VL
i 0 ' X-v
R-----2:
Y'
Then,
(i) Impedance, Z = ~R2 + xi = Vnns
Inns
=~R2+ro2L2 [':XL=roL]
(ii) For the phase angle, tan <1> = XL = roL ,R R
voltage leads current by phase <1>.
(iii) If V = Vo sin rot, then 1= 10 sin (rot - <1»
A R-C Series AC CircuitR e
-1 r:-1
ACsource(Circuit diagram)
y
VROt.:-r:--------"---XIP :
.: Voltage lags current by phase IPII
Vc ---------- 'Vy' Phasor diagram
Chapterwise eSSE Solved Papers PHYSICS
Then,
(i) Impedance, Z = Vrms = ~R2 + X~Inns
= ~R2 + ro;c2 [.: x; = ~]
(ii) For the phase angle, tan <1> = Xc = _1_R roCR
(iii) If V = Vo sin rot, then I = 10 sin (rot + <1»
. f R R(IV) Power actor, cos <1> = - = r====Z ~R2 + X~
An L - C Series AC Circuit(i) Impedance, Z = Vnns = XL - Xc
Inns
(ii) Applied voltage = VL - Vc
(iii) Phase difference between voltage and current is~,2
(iv) Power factor, cos <1> = O,
(v) Current, I = IOSin( rot±~)
An L- C- R Series AC Circuit-VR I VL--t<-Vc---..
R L e
VL r~I FI
AC sourceVc~------~'"'-'r---~-----
Circuit diagram
~ VL
1B'! ,VVL :
I07tl2 IP :t 1t/2 A
Vc -VR-
!e
.... ·XI
: VcY'
Phasor diagram
CHAPTER 7 : Alternating Current
Then,
(i) Impedance, Z = ~R2 + (XL - xci = Vnns
i.;
= R2 + (wL- w~r(ii) If XL> Xc' then V leads I by <I> and if
XL < Xc' then V lags behind I by <1>.
X -X V-v:where, tan <I> = L C = _L__ C
R VR
1 1wL-- 21tvL---_---'w=C"'- = 21tVC
R R(iii) If net reactance is inductive circuit behaves
as L-R circuit.(iv) If net reactance is capacitive, circuit
behaves as C-R circuit.
Resonant L-C -R Series AC Circuit(i) XL = x;
(ii) Impedance, Z = Zmin = R i.e. circuitbehaves as resistive circuit.
(iii) The phase difference between V and Iis 0°.
(iv) Resonant angular frequency,1 1
000 = &::::} Vo = 21t& Hz
(v) Average power consumption Pay becomesmaximum.
(vi) Current becomes maximum and
I = Vrmsmax R
Quality FactorThe characteristic of a series resonant circuit isdetermined by Q-factor. It indicates thesharpness of resonance in an L -C -R series ACcircuit.
. V v: wL 1 IJ¥Quality factor = .....k.= ~ = _0_ = -- = - -VR VR R 00 oCR R C
Q = 000
002 - 001
209
where, 001 and 002 are the frequencies when currentdecreases to 0.707 times of the peak value ofcurrent.Quality factor is also defined as
Q=21t( Maximum energy stored )Energy dissipated per cycle
L-C OscillationsConsider the L-C circuit shown as below:
+jC_+ -+ -CJ+jC:+ -
1=0 L - Imax
Completelyelectricalenergy
"" ""I " ,. I' .. ::::-.....:: .. 'Completely
magnetic energy
When the charged capacitor is connected with theinductor, current flows through the inductor andenergy stored in the inductor in the form ofmagnetic field and capacitor discharges andvice-versa. In this way, energy oscillates betweencapacitor and inductor. If the circuit is not idealoscillations finally die away.
The frequency of oscillation is
000 = -k ::::}fo = ~~ = 21tk
PREVIOUS YEARS'EXAMINATION QUESTIONSTOPIC 1o 1 Mark Questions
1. Define 'quality factor' of resonance inseries L-C-R circuit. What is its 81 unit?Delhi 2016
2. The power factor of an AC circuit is 0.5.What is the phase difference betweenvoltage and current the circuit? Foreign 2016
210
3. Define capacitor reactance. Write its 81units. All Indio 2015
4. Plot a graph showing variation ofcapacitive reactance with the change inthe frequency of the AC source.All Indio 2015C
5. Why is the use of AC voltage preferredover DC voltage? Give two reasons.All Indio 2014
6. When an AC source is connected across aninductor, show on a graph the nature ofvariation of the voltage and the currentover one complete cycle.Delhi 2012C
7. The peak value of emf in AC is Eo. Writeits (i) rms (ii) average value over acomplete cycle. Foreign 2011
8. The current flowing through a pureinductance 2mH is, 1= (15 cos 300t)A. Whatis the (i) rms and (ii) average value ofcurrent for a complete cycle? Foreign 2011
9. Define the term wattless current. Delhi 2011
10. A reactive element in an AC circuit causes. the current flowing
(i) to lead in phase by reI 2(ii) to lag in phase by nl 2
w.r.t. the applied voltage. Identify theelement in each case. Deihl 2010C
11. Define the term rms value of the current.How is it related to the peak value?All Indio 2010C
12. How much average power over a completecycle, does an AC source supply to acapacitor? Delhl2009C
13. An AC current, 1 = 10 sin rot producescertain heat H in a resistor R over a timeT = 2re/ro. Write the value of the DCcurrent that would produce the same heatin the same resistor in the same time.All Indio 2DD9C
Chapterwise C8SE Solved Papers PHYSICS
o 2 Marks Questions14. A capacitor C, a variable resistor R and a
bulb B are connected in series to the ACmains in the circuit as shown. The bulbglows with some brightness. How will theglow of the bulb change if
(i) a dielectric slab is introducedbetween the plates of the capacitorkeeping resistance R to be the same
(ii) the resistance R is increased keepingthe same capacitance? Deihl 2014
15. The figure shows a series L-C-R circuitconnected to a variable frequency 250 Vsource with L = 50 mH, C= 80J..l.FandR=40n.
Determine(i) the source frequency which derives
the circuit in resonance.(ii) the quality factor (Q) of the circuit.
All Indio 2D14C
16. The figure shows a series L-C-R circuitconnected to a variable frequency 250 Vsource with L = 40mH, C = 100 JlF andR=50n.
Determine(i) the source frequency which derives
the circuit in resonance,(ii) the quality factor (Q) of the circuit.
All Indio 2014
CHAPTER 7 : Alternating Current
17. The figure shows a series L-C-R circuitconnected to a variable frequency 220 Vsource with L = 80 mH,C = 50 IlF andR= 60Q.Determine
(i) the source frequency which derivesthe circuit in resonance.
(ii) the quality factor Q of the circuit.All Indio 2014
18. Show that the current leads the voltagein phase by reI 2 in an AC circuitcontaining an ideal capacitor. Foreign 2014
19. In a series L-C-R circuit, obtain theconditions under which
(i) the impedance of circuit is minimumand
(ii) wattless current flows in the circuit.Foreign 2014
20. A series L-C-R circuit is connected to anAC source (200 V, 50 Hz). The voltagesacross the resistor, capacitor andinductor are respectively 200 V, 250 Vand 250 V.
(i) The algebraic sum of the voltagesacross the three elements is greaterthan the voltage of the source. Howis this paradox resolved?
(ii) Given the value of the resistance ofR is 40 Q, calculate the current inthe circuit. Foreign 2013
21. A resistor 'R' and an element 'X' areconnected in series to an AC source ofvoltage. The voltage is found to lead thecurrent in phase by rt I 4. If 'X is replacedby another element 'Y, the voltage lagsbehind the current by rt I 4.
(i) Identify elements 'X and 'Yo(ii) When both 'X and 'Yare connected in
series with 'R' to the same source,will the power dissipated in the circuitbe maximum or minimum? Justifyyour answer. Foreign 2013
211
22. Write the expression for the impedanceoffered by the series combination ofresistor, inductor and capacitor connectedto an AC source of voltage V = Vo sin oz.Show on a graph the variation of thevoltage and the current with '(Of in thecircuit. All Indio 2012C
23. A lamp is connected in series with acapacitor. Predict your observation whenthis combination is connected in turn across
(i) AC source and(ii) a DC battery. What change would your
notice in each case if the capacitance ofthe capacitor is increased? Delhi 2012C
24. Calculate the quality factor of a seriesL-C-R circuit with L = 2.0 H, C = 21lF andR = 10 Q. Mention the significance ofquality factor in L-C-R circuit. Foreign 2012
25. An alternating voltage given byV = 140 sin 314t is connected across a pureresistor of 50 Q. Find
(i) the frequency of the source.(ii) the rms current through the resistor.
All Indio 2012
26. An alternating voltage given byV = 280 sin 50 tu is connected across a pureresistor of 40 Q. Find
(i) the frequency of the source.(ii) the rms current through the resistor.
All Indio 2012
27. An alternating voltage given byV = 70 sin 100 ret is connected across a pureresistor of 25 Q. Find
(i) the frequency of the source.(ii) the rms current through the resistor.
All Indio 2012
28. A light bulb is rated 150 W for 220 V ACsupply of 60 Hz. Calculate
(i) the resistance of the bulb(ii) the rms current through the bulb.
All Indio 2012
29. Prove that an ideal capacitor in an ACcircuit does not dissipate power.Delhi 2011
212~.... _._ .._-- .. ~.--
30. An electric lamp having coil of negligibleinductance connected in series with acapacitor and an AC source is glowing withcertain brightness. How does thebrightness of the lamp change on reducingthe (i) capacitance, and (ii) the frequency?Justify your answer.
tjSource Deihl 2010
31. (i) An alternating voltage V = Vm sin rot·a.pplied to a series L-C-R circuitderives a current given byI = 1m sin (ro t + <1». Deduce anexpression for the average powerdissipated over a cycle,
(ii) For circuit used for transportingelectric power, a low power factorimplies large power loss intransmission. Explain.Foreign 2011; Deihl 2009
32.. (i) The graphs (I) and (II) represent thevariation of the opposition offered bythe circuit element to the flow ofalternating current with frequencyof the applied emf. Identify thecircuit element corresponding to eachgraph.
(ii) Write the expression for theimpedance offered by the seriescombination of the above twoelements connected across the ACsources.Which will be ahead in phase in thiscircuit, voltage or current?
c.Q C 1--------.~ Q)
(I) ~ t:::8:50.9
o Frequency --
ehopterwise eSSE Solved Papers PHYSICS
c.Q C
(II) 'lij ~8.:Sa.()0.9
Frequency --All Indio 20lle
o 3 Marks Questions33. (i) Find the value of the phase
difference between the current andthe voltage in the series L-C-Rcircuit shown below. Which oneleads in phase: current or voltage?
V=Vo sin (1000 t + CP)
(ii) Without making any other change,find the value of the additionalcapacitor C', to be connected inparallel with the capacitor C, inorder to make the power factor of thecircuit unity. Delhi 2017
34. (i) When an AC source is connected toan ideal inductor show that theaverage power supplied by thesource over a complete cycle is zero.
(ii) A lamp is connected in series withan inductor and an AC source. Whathappens to the brightness of thelamp when the key is plugged in andan iron rod is inserted inside theinductor? Explain.
Lamp
All Indio 2016
CHAPTER 7 : Alternating current
35. Calculate the value of the additionalcapacitor which may be joined suitably tothe capacitor C that would make thepower factor of the circuit unity.
R=400 n
V,V,,'o (1000I+;) +_~:r---f~'~L=100mH
All India 2015
36. A circuit containing an 80 mH inductorand a 250 mF capacitor in seriesconnected to a 240 V, 100 radls supply.The resistance of the circuit is negligible.
(i) Obtain rms value of current.(ii) What is the total average power
consumed by the circuit? Deihl 2015C
37. A source of AC voltage V = Va sinroz isconnected to a series combination of aresistor 'R' and a capacitor 'C'. Draw thephasor diagram and use it to obtain theexpression for (i) impedance of the circuitand (ii) phase angle. All India 2D15C
38. (i) Determine the value of phasedifference between the current andthe voltage in the given series L-C-Rcircuit.
R=400n
V=Vosin(1000t+CP) "-'
L=100 ~H(ii) Calculate the value of additional
capacitor which may be joinedsuitably to the capacitor C thatwould make the power factor of thecircuit unity. Delhi 2015
39. An inductor L of inductance XL isconnected in series with a bulb B and anAC source. How would brightness of thebulb change when (i) number of turn inthe inductor is reduced, (ii) an iron rod isinserted in the inductor and (iii) acapacitor of reactance Xc = XL isinserted in series in the circuit. Justifyyour answer in each case. All India 2015
213
40. A voltage V = Va sin rof is applied to aseries L-C-R circuit. Derive the expressionfor the average power dissipated over acycle.Under what condition is (i) no powerdissipated even though the current flowsthrough the circuit (Ii) maximum powerdissipated in the circuit? All India 2014
41. In a series L-C-R circuit connected to anAC source of variable frequency andvoltage V = Vm sin rot, draw a a plotshowing the variation of current (1) withangular frequency (oi) for two differentvalues of resistance, R,. and R2(R,. > R2).Write the condition under which thephenomenon of resonance occurs. Forwhich value of the resistance out of the twocurves, a sharper resonance is produced?Define Q-factor of the circuit and give itssignificance. Deihl 2013
42. (i) For a given AC, i = im sin rot, showthat the average power dissipated in aresistor R over a complete cycle is1 '2R-~m .2
(ii) A light bulb is rated at 100 W for a220 V AC supply. Calculate theresistance of the bulb. All India 2013
43. (i) When an AC source is connected to anideal capacitor, then show that theaverage power supplied by the sourceover a complete cycle is zero.
(ii) A lamp is connected in series with acapacitor. Predict your observationswhen the system is connected firstacross a DC and then an AC source.What happens in each case if thecapacitance of the capacitor isreduced? Delhi 2013C
44. The figure shows a series L-C-R circuitwith L = 10 H, C = 40IlF, R = 60 Qconnected to a variable frequency 240 Vsource. Calculate
214
(i) the angular frequency of the sourcewhich derives the circuit atresonance.
(ii) the current at the resonatingfrequency.
(iii) the rms potential drop across theinductor at resonance. Delhi 2012
45. A series L-C-R circuit is connected to anAC source. Using the phasor diagram,derive the expression for the impedanceof the circuit. Plot a graph to show thevariation of current with frequency of thesource, explaining the nature of itsvariation. All India 2012
46. A series L-C-R circuit is connected to a220 V variable frequency AC supply. IfL = 20mH, C = (800l1t2) IlF and R = 110 n.
(i) Find the frequency of the source forwhich average power absorbed bythe circuit is maximum.
(ii) Calculate the value of maximumcurrent amplitude. Deihi 2010C
47. An AC voltage V = Va sin cot is appliedacross a pure inductor L. Obtain anexpression for the current I in the circuitand hence obtain the
(i) inductive reactance of the circuitand
(ii) the phase of the current flowingwith respect to the applied voltage.All India 2010C
48. An AC voltage, V = Va sin cot is appliedacross a pure capacitor, C. Obtain anexpression for the current I in the circuitand hence obtain the
(i) capacitive reactance of the circuitand
(ii) the phase of the current flowingwith respect to the applied voltage.All India 2010C
49. The graphs shown here depict thevariation of current Irms with angularfrequency ro for two different series
Chopterwise CBSE Solved Papers PHYSICS
L-C-R circuits.1.0
1t+ (I)
I ~ (II)
VJ ,'/ .........- 0
~E 0.5....•
1o 0.5 1.0 1.5 2.0
ro(rad/s) ---
Observe the graphs carefully.(i) State the relation between Land C
values of the two circuits when thecurrent in the two circuits ismaximum.
(ii) Indicate the circuit for which(a) power factor is higher(b) quality factor Q is larger. Give thereasons for each case. Deihl 2009C
o 5 Marks Questions50. A device X is connected to an AC source,
V = Va sin wt. The variation of voltage,current and power in one cycle is shown inthe following graph.
Ay
I
C 1t l__ I
- III
(i) Identify the device X.(ii) Which of the curves A, Band C
represent the voltage, current and thepower consumed in the circuit? Justifythe answer.
(iii) How does its impedance vary withfrequency of the AC source? Showgraphically.
(iv) Obtain an expression for the currentin the circuit and its phase relationwith AC voltage. All India 2017
CHAPTER 7 : Alternating current
-5'f: An AC source of voltage V = Va sin rot isconnected to a series combination of L, Cand R. Use the phasor diagram to obtainexpressions for impedance of the circuitand phase angle between voltage andcurrent. Find the condition when currentwill be in phase with the voltage. What isthe circuit in the condition called?Oelhl2016
52. A 2/lF capacitor, 100 Q resistor and 8 Hinductor are connected in series with anAC source.
(i) What should be the frequency of thesource such that current drawn inthe circuit is maximum? What is thisfrequency called?
(ii) If the peak value of emf of the sourceis 200 V, find the maximum current.
(iii) Draw a graph showing variation ofamplitude of circuit current withchanging frequency of appliedvoltage in a series L-C-R circuit fortwo different values of resistance Rtand R2(~ > R2).
(iv) Define the term 'Sharpness ofResonance'. Under what condition,does a circuit become more selective?Foreign 2016
53. (i) An AC source of voltage V = Va sin rotis connected across a seriescombination of an inductor, acapacitor and resistor. Use thephasor diagram to obtain theexpression for(a) impedance of the circuit and(b) phase angle between the voltage
and the current.(ii) A capacitor of unknown capacitance,
a resistor of 100 Q and an inductorof self inductance L = (4/ 7t2) henryare in series connected to an ACsource of 200 V and 50 Hz. Calculatethe value of the capacitance and thecurrent that flows in the circuitwhen the current is in phase withthe voltage. All India 2013C
215
54. (i) A series L-C-R circuit is connected to anAC source of variable frequency. Drawa suitable phasor diagram to deducethe expressions for the amplitude of thecurrent and phase angle.
(ii) Obtain the condition at resonance.Draw a plot showing the variation ofcurrent with the frequency of ACsource for two resistances R, andRiRl > R2). Hence, define the qualityfactor Q and write its role in thetuning of the circuit. Delhl2014C
55. A voltage V = Va sin rot applied to a seriesL-C-R circuit drives a current i = io sin rot inthe circuit. Deduce the expression for theaverage power dissipated in the circuit.For circuits used for transporting electricpower, a low power factor implies largepower loss in transmission. Explain. Definethe term 'Wattless current'. Delhl2012C
56. Derive an expression for the impedance ofa series L-C-R circuit connected to an ACsupply of variable frequency. Plot a graphshowing variation of current with thefrequency of the applied voltage. Explainbriefly how the phenomenon of resonancein the circuit can be used in the tuningmechanism of a radio or a TV set? Delhi 2011
57. (i) What do you understand bysharpness of resonance in a seriesL-C-R circuit? Derive an expressionfor Q-factor of the circuit.
(ii) Three electrical circuits having ACsources of variable frequency are shownin the figures. Initially, the currentflowing in each of these is same. If thefrequency of the applied AC source isincreased, how will the current flowingin these circuits be affected? Give thereason for your answer.
CJQwLEE
HOTS; Deihl 2011C
216
58. A series L-C-R circuit is connected to anAC source having voltage V = Vm sin rot.Derive the expression for theinstantaneous current I and its phaserelationship to the applied voltage.Obtain the condition for resonance tooccur. Define power factor. State theconditions under which it is
(i) maximum and(ii) minimum. All India 2010
59. A resistor of 400 n, an inductor of ~ HandII I.. 50 1t
a capacitor of - JlF are connected in series1t
across a source of alternating voltage of140 sin 100 ta V. Find the voltage (rms)across the resistor, the inductor and thecapacitor. Is the algebraic sum of thesevoltages more than the source voltage? Ifyes, resolve the paradox.(Given, .J2 = 1.414). Foreign 2010
60. (i) Derive an expression for the averagepower consumed in a series L-C-Rcircuit connected to AC source forwhich the phase difference betweenthe voltage and the current in thecircuit is <\> •
(ii) Define the quality factor in an ACcircuit. Why should the qualityfactor have high value in receivingcircuits'? Name the factors on which it
. depends. Delhi 2009
61.i An AC source of emf, E = Eo sin rot isIcdnflected across a series combination ofan inductor L, a capacitor C and a resistor,R. Obtain an expression for the equivalentimpedance Z of the circuit and hence, findthe value of rofor the AC source for whichZ=R.
Show that the phase angle <\> (between thecurrent flowing in this circuit and thevoltage applied to it) can be obtainedthrough the relation.
<\> = tan'" (Reactive. impedance)Resistance
All India 2009
o ehapterwise eBSE Solved Papers PHYSICS
o Explanations1. The quality factor (Q) of resonance in series
L-C-R circuit is defined as the ratio of voltagedrop across inductor (or capacitor) to the appliedvoltage, (1/2)
i.e, Q= VL = 10XL = rooL = _1_VR loR R rooCR
It is an indicator of sharpness of the resonance.Quality factor has no unit. (112)
2. Power factor, coso = 0.5cos Ij) = cos 60° => Ij) = 60°
=> Ij) = 60°Phase difference = 60° (1)
3. Capacitor reactance is the resistance offered by acapacitor, when it is connected to an electric. '1" bX 1 IC1rcUIt. t 1Sglven y c = - = --
roC 21tjC (112)
where, ro = angular frequency of the sourceC = capacitance of the capacitor
Instantaneous value of alternating currentthrough a capacitor,
1 = ~sin(rot+ 1t/2)I I roC
= 10Sin( rot + ~)
On comparing with Ohm's lawE10=--
l/roCThe SI unit of capacitor reactance is Ohm (Q).
(112)
4. Showing variation of capacitive reactance withthe change in the frequency of the AC sourcewith graph is given below.
XC=~C=2!C tLXcoc ~ Xc
v a v---+ (1)
5. The use of AC voltage ispreferred over DC voltage because of thefollowing reasons:(i) The loss of energy in transmitting, the AC
voltage over long distances with the help ofstep-up transformers is negligible ascompared to DC voltage. (112)
(ii) AC voltage can be stepped up and steppeddown as per the requirement by using atransformer. (112)
CHAPTER 7 : Alternating Current
6. v = Vo sinter
I = IOSin( oot-~)(1/2)
(1/2)
7. Eo = peak value of emf in a complete cycle,
(i) rms value [Enns] = lz(ii) average value [E) = zero
(1/2)
(1/2)
8.-,
To calculate different values for AC with the help of 'I
given equation, compare the given equation withstandard equation of AC. I-- ._._. . J
Current flowing through the inductor,I = 15 cos 300t
Comparing with I = Io sin ootHere, peak value of current,
Io =15 A
(i) For complete cycle, rms value of current
I =~=~Anns -J2 -J2
(ii) For complete cycle, average value of current iszeroi.e. (112 x 2 = 1)
9. Wattless Current The current in an AC circuitwhen average power consumption in AC circuitIszero, is referred as wattless current.If <I> is the phase difference between voltage andcurrent, then power associated with I sin <I>
component of current is termed as wattlesscurrent. (1)
10. (i) In case of pure capacitive circuit, the currentleads in phase by 7t/2with respect to theapplied voltage. $0, the element will be acapacitor. (1/2)
(ii) In case of pure inductive circuit, the currentlags in phase by 7t/2with respect to theapplied voltage. $0, the element will be aninductor. (112)
11. It is defined as the value of Alternating Current(AC) over a complete cycle which would generatesame amount of heat in a given resistor that isgenerated by steady current in the same resistor
217
12.
and in the same time during a complete cycle.It isalso called virtual value or effective value of AC.Let the peak value of the current be Io
I - Io _ Io• • nns - -J2 ~ Irms - -J2where, Io = peak value of AC.Average power, Pay = Vnns X Irms X cos <I>
(1)
But for pure capacitive circuit, <I> = 90° = .::2
Pay = Vrms X Irms X cos 90° = 0
Pay = 0 (1)
13. An AC current I = Io sinmr produces certain heatH in a resistor R over a time T = 2x 3.14/00, is
given by H = (~rRT and the same amount of
heat produced by DC current in same time isgiven by H = I2RT. As these heats are equal, then
I2RT= (~)RTI
$0, I = Jx where I stands for DC and Io is the
peak value of AC current.(1)
(i) As, the dielectric slab is introduced betweenthe plates of the capacitor, its capacitance willincrease. Hence, the potential drop across thecapacitor will decrease, i.e. V = g As a result,
Cthe potential drop across the bulb will increaseas they are connected in series.Thus, itsbrightness will increase. (1)
(ii) As the resistance R is increased, the potentialdrop across the resistor will increase. As aresult, the potential drop across the bulb willdecrease as they are connected in series. Thus,its brightness will decrease. . (1)
Given, L = 50mH = 50xlO-3HC = 80ll-F = 80 X 1O-6FR = 40(20 V = 200V
(i) In the L-C-R, the resonant angular frequency isgiven by
1 1000 = -- = 500 rad /s.JLC ,,)50 X 10 3 X 80 xl0-6
:. The actual/source frequency is given by
00= 27tv ~ v= ~27t
14.
15.
~ v = 500 = 250 = 79.61 '" 80Hz~ 7t (1~
218
(ii) Quality factor, Q = oooL = 500 x 50 X 10-3
= 0.625R 40
c 'c (112)
~{ J,%~ei'to Ans. 15. (2)
17. Refer to Ans. 15. (2)
18. Let us consider a capacitor C connected to an ACsource as shown below:
- +cLet the AC voltage applied by
V =Vmsinoot
v=iC
Applying Kirchhoff's loop rule, we have
Vm sinoot = i ~q = CVm sinootC
Al . dq . d (V.' )so, I = - ~ I = - m smootdt dt
i = ooCVm cosootWe know that cosO>t= sineor + rc/2In the circuit,
V. . X . Im = 1m c=lm ooC
~ im = VmooC ... (ill)Substituting the values of Eqs: (ii) and (ill) inEq. (I), we get e- _', '
... (i)
... (ii)(1)
. . . ( rc)l=lmsm oot+- .2 ,I
The phase diagram which shows the current lead(S rfhe voltage in phase by 90° is given below:
Vf'I ~i~O>t_i, 1'------:.-- --- ------- ----- -- -------- - ~
I "(S\f) : \r~' O~O>t7T,~rcr-~-+-~~\=-
~~~+~ ~ \
V
19. (i) The impedance of a series L-C-R circuit is
given by Z = IR2 + (OOL _ ~)2.. , V ooC
Z will be minimum when ooL= l.-, i.e. whenooC
the circuit is under resonance. Hence, in thiscondition Z will be minimum and equal to R.
(1)
o ehopterwise eBSE Solved Papers PHYSICS
(ii) Average power dissipated through a seriesL-C-R circuit is given by
Pay = Eyly cos owhere, Ey =rrns value of alternating voltage
Iv = rms value of alternating current<I> = phase difference between
current and voltageFor wattless current, the power dissipatedthrough the circuit should be zero.
i.e. coso = 0 ~ cos o =cos~ ~ <I> = ~2 2
Hence, the condition for wattless current isthat the phase difference between the currentand the circuit is purely inductive or purelycapacitive in which the voltage and currentdiffer by a phase angle of 1[/2, i.e. <I> = ± ~
2 (1)
20. (i) From given parameter VR = 200V, VL = 250Vand Vc = 250 V. VeCC should be given as
Veff = VR + VL + Vc= 200V + 250V + 250V = 700V
r::J(250 V, 50 Hz) (1)
However, Verc > 200V of the AC source. Thisparadox can be solved only by using phasordiagram, as given below
(VeCC)= ~vi+ (VL - Vel2
(1)
Since, VL = VcSo, VeCC= VR = 200V
(ii) Given, R = 40n, so current in the L-C-R circuit.(1)
I rr = VeCC= 200 = 5Ae R 40 [XL = Xc or Z = R]
21. (i) In R-L series combination, voltage leads thecurrent by phase <I> = ~. It means element X is
4an inductor (with reactance equal to R). In R-Cseries combination, voltage lags behind the
CHAPTER 7 : Alternating Current
current by phase ~ = ~. So, element Y is a4
capacitor (with reactance equal to R).(ii) If both elements X and Yare connected in
series with R. then power dissipation in thecombination can be given asP = Vrms ·lrms· coso
R Rcos ~ = - = ---r=~===~
Z ~R2 + (XL - Xcl2
(1)CX~d]Here, XL = Xc = R.So, cos o = 1
Hence, P = Vrms1rms(Maximum)
22. Impedance offered by series L-C-R circuit,
Z = ~R2 + (Xc - Xrl2, V = ~vi+ (VC - VL)2.
As Vc and VL are the voltages applied acrosscapacitor C and inductor L. Vc or VL may begreater than n V.The situation may be shown in figure, where Vc > V.
23. (i) On increasing capacitance, current willincrease. It also increases the brightness ofbulb. As capacitance increases, capacitivereactance (Xc = 1/ roC) decreases,impedance Z decreases, hence currentincreases. (1)
(ii) There will no flow of current and hence bulbwill not glow. (1)
24. Given, L = 2.0 H
C = 21lF = 2 x 10- 6 F
R=10n
Q_ factor = ~ fI = ~ ~ 2RVC 10 2 X 10-6Now,
10 x 10
= _1_ =10010-2
Quality factor is also defined as
219
(1)
Q = 21tf x Energy storedPower loss
So, higher the value of Q means the energy loss isat lower rate relative to the energy stored, i.e. tll!oscillations will die slowly and damping would beless. (1)
25. Given, V = 140 sin 314t, R = 50 nComparing it with V = Vo sin rot(i) Here, ro = 314 rad/s
i.e. 21tV= 314314
::;> v=-21t
(112)
[':ro = 21tv]
) .
(1)
314v=---zx 3.14
= 50HzFrequency of AC, v = 50 Hz
( .. ) A I - Vrms d V. - Vo11 S, rms - R an rms - .J2
Here, Vo = 140 V
::;> V. = 140 x .J2 = 70.J2 Vrms .J2 .J2
I = 70.J2 = 70./2•. nns R 50
(1/2)
(112)
(1)
=1.9Aor2A
26. Refer to Ans. 25.
27. Refer to Ans. 25.
(112)
(Ans.25Hz,4.95A)
(Ans. 50Hz, 1.98 A)
28. (i) P = 150W, V = 220Vv2
Resistance of the bulb, R = -P (1/2)
R = 220 x 220 = 322.7 n150 - (1/2)
(ii) As I = Vrms = ~ (Vrms = V = 220V)'rms R 322.7
::;> I rms = 0.68A(1/2)
(1/2)
29. Since, average power consumption in an ACcircuit is given by
Pav = Vrms x lrms x cos ~ (1)
But in pure capacitive circuit, phase differencebetween voltage and current is given by
(1)
220 o Chapterwise cast Solved Papers PHYSICS
P v: x Inns X cos 2:av= rms 2
Pay = 0 C cos ~ = 0)
Thus, no power is consumed in pure capacitiveAC circuit. (1)
30. When AC source is connected, the capacitorff . . XIIo ers capacinve reactance c = - = --.
roC 27tVoCThe current flows in the circuit and the lampglows. (1)
(i) On reducing capacitance C, Xc increases socurrent in the circuit reduces. Therefore, thebrightness of the bulb reduces. (1)
(ii) On reducing frequency v, Xc increases socurrent in the circuit reduces. Therefore, thebrightness of the bulb reduces.
31. (i) Let at any instant, the current and voltage inan L-C-R series AC circuit is given by
1=10 sin cor
V = Vo sin (wt + cp)The instantaneous power is given by
P = VI = Vo sin (wt + cp) 10 sin wt
P= Vo 10 [2sinwtsin(wt+ cp)]2
V.IP = VI =.-..LQ. [cos cp - cos (2wt + cp)] ... (i)
2
[':2 sin A sin B = cos (A - B) - cos (A + B)]
Work done for a very small time interval dt isgiven by
dW = Pdt
dW = Vldt:. Total work done over T, a complete cycle isgiven by
W = iT Vldt~ - ..:; o· (112)
W (VldtBut P =_=_0 __
av T T
1 iT~ P = - Vldtav T 0
1 T V. I= - i .-..LQ. [cos cp - cos (2wt + cp)] dtT 0 2
V./[T T ]or Pay = ~o facos cp dt - fo cos (2wt + cp) dt
= Volo [cos cp [t]~ - 0] (By trigonometry)2T
V. I V. Ior P = ~ x cos cp x T = ~ x cos cp
av 2T 2
_ Vo 10 n.Pay - ..J2 x ..J2 cos 'I'
32.
~ Pay = Vnns x Irms x cos cpThis is the required expression.
(ii)': Power factor, cos cp = ~Z
where, R = resistance and Z = impedanceLow power factor (cos cp) implies lower ohmicresistance and higher power loss as Pay oc .!.. in
Rpower system (transmission line) (1)
(i) From graph (I), it is clear that resistance(opposition to current) is not changing withfrequency, i.e. resistance does not depend onfrequency of applied source, so the circuitelement here is pure resistance (R).
From graph (II), it is clear that resistanceincreases linearly with frequency, so thecircuit element here is an inductor.Inductive resistance, XL = 21tfl.
~ XL oc f (1)
(ii) Impedance offered by the series combinationof resistance R and inductor L.
Z = ~R2 + xi = ~R2 + (21tfl.)2
In L-R circuit, the applied voltage leads thecurrent in phase by 2:.
2 n)Given, L = 100 mR = 100 x 1O-3R,
C=2J!F = 2xl0-6F, w=1000andR=400n
(1/2)
33.
(i) For phase difference,
(WL __ 1 )
tan cp = wCR
[where, cp is phase differencebetween current and voltage]
.: 00 = 1000 ~wL = 1000 x 100 X 10-3 = 100 n(112)
1 = __ 1_= 500 nwC 1000 x 2 X 10-6 2 X 10-3
tan cp = 100 - 500 = -400 = _ 1400 400
cp = tan"! (-1) ~ cp = 135°
1wL < - or XL < Xc
wCSince,
CHAPTER 7 : Alternating Current
Therefore, current is leading in phase by aphase angle 135°. (1/2)
(ii] For unit power factor cas $ = 1~ R =1
R2 + (WL __ 1 )2wC'
where, C'is the total capacitance.
~ R2 + (WL __ 1_)2 = R2«c:
wL =_1_wC'
wL = 100 = ....!...- = _1_ C'oc 1000
C' = _1_ = 10-5 F = 10 IlF105
Additional capacitance C'required in parallel= C' - C = 10 11F - 211F = 8 11F (112)
34. (i) As Pay = v,m.Irms cosoIn ideal inductor, current Irms lags behind appliedvoltage Vrms by 1t I 2.:. $ = 1t/2 so, Pay = v,m.Irms cos 1t/2or Pay = Vrm.Irms X O.or Pay = 0 (1)
(ii) Brightness of the lamp decreases. It is becausewhen iron rod is inserted inside the inductor,its inductance L increases, thereby increasingits inductive reactance XL and henceimpedance Z of the circuit. As Irms = Vrms, so,
Zthis decreases the current Irms in the circuitand hence the brightness of lamp. (2)
35. For unity power factor, XL = Xc
wL = _1_ [':C' = C+ C"1we'
C,=_l_w2L
C' = C+ C"C" = C' - C = 10 - 2 = 81lF
So, required capacitor is 81lF which is added inparallel with the given capacitor.
36. (i) Here, L = 80mH, C= 250mH,w=100rad/sVrms = 240V
Reactance = 1wL - ~ 1
=II00X80XlO-3 1 1100 x 250 X10-3
221
(112)
=18- ;51=7.96
I = Vrmsrms Reactance
= 240 = 30.15A7.96
(ii) The total average power consumed by circuitzero. (1)
37. (i) V = Vosinwt " ... (i)From diagram, by parallelogram law of vectoraddition, VR + Vc = V (1)Using pythagorean theorem,
Y VR..•.••......• V
Vcwe getV2 = Vi + vl = (IR)2 + (LXd2
V2 = I2(R2 + X~)
I= V =.!'::JR2+ X~ Z
~ ~2 Iwhere,Z=VR +Xc= R +2"2wC
Z = impedance.(ii) The phase angle $between resultant voltage
and current is given by (2)
tane = Vc = LXc = Xc = I/wC = _1_VR IR R R wRC
38. (i) Consider the given figure,
R=400Q
V=Vosin(1000t+<I» "--
L=100 JlHSince, the alternating emf in the above L-C-R seriescircuit would be represented by
V= Vosin(lOOOt+ $) ~ w=1000HzGiven, R = 400 Q, C = 21lF, L = 100 mH
. . I': Capacitive reactance, Xc = -we
(2) X _ 1c - 1000 x 2 x 1O~
222
.: Inductive reactance. x, -z-: wL~ XL = 1000 x 100 X 10-3 ~ XI.:= loonSo, we can see that Xc > X[~ tan $ is negative.Hence. the voltage lags behind the current hy .I
phase angle $. The AC circuit is capacuauccdominated circuit..: Phase difference, tan $ = XL=.~~
100 - 500 R-400tan o = --- => tan o = .----, t.an$ ",-1400 400
~ tan$=-tan(-~) => ¢= __ IT4 ·l
This is the required value of the phase differencebetween the current and the voltage in the givenseries L-C-R circuit. (1)As, cas $ < 1.
(ii) Suppose, new capacitance of the circuit is C' .Thus. to have power factor unit)'
cas $' = I ~ --:;R2:-'7i--=--;;;)2~ Rl = R2 + (XI, _ X~,)2
~ XL = Xc = _L_ or wL =«c wC'2 I 1 I=> 00 =- or (I 000) = ----u: LC'
(':w = 1000)
39.
~ C'= __ I_ = J __. _t.« 106 100 X JO-J X 10"
= ~~ = ~_ = 10-5
106 IOS
~ C' = 10-5 F = lOx 10-6 F = I0~lF
As, C' > C. Hence. we have to add an additionalcapacitor of capacitance SJ.LF (IOJ.LF-2J.LF) inparallel with previous capacitor. (1)
The inductive reactance (X I.) is XL = wL
where, w = angular frequency of AC source,L = inductance of the inductor.
I+--VR=RJ II. Vc=XcJ-1
R
£0 sin rotSeries L-R circuit
o Chopterwise CBSE Solved Papers PHYSICS
(1)
The net resistance 01 the circuit is given by
J 2 2Z =. XL + R
where, f{ = Icsistancc or Ihe bull>(i) We known that if the number of turns in the
inductor decreases. then inductance Ldecreases, So. Ihe net rcsistaru « of the circuitdecreases and hence, the current through thecircuit increases. iru rcasinj; the brightness ofthe bulb. (1)
(Ii) If the soft iron rod h inserted ii: rue inductor.then the induc.t.mcc L inrrcasc«. Therefore, thecurrent through the bulb will decrease.dccrcasinu the ilrightllv,;s of the bulb.
I = frn~rms Xl
(1)
(iii) If the capacitor of reactance Xc =X[ isconnected in series with the circuit, then
Z = ~(XI - Xd2 + R2
~ Z=R 1':XL=XclThis is a case of resonance. In this case, themaximum current will flow through thecircuit. Hence, the brightness of the bulb willincrease. (1)
40. Let applied alternating voltage,V = Vosinwt
We know that,J = J 0 sin (m! - $)
Power, P = VI
... (i)
... (ii)(1)
Substituting the value of V and 1 from Eq. (i)and Eq. (ii), we getP = Vosinwt,Jo sin (wi - $)So, the instantaneous power is given by':sin (wi - $) = sin wtcos$ - cas wt sin $(i) If $ = 90°, then no power is dissipated even
though the current flows through the circuit.Pay =0
tan$=( XL ~ Xc)wL--L
wCtan e = (,: tan 90° = 00)R (1)
(ii) If $ = 0°, then maximum power is dissipated inthe circuit.
Pay = maximumwL _ 1_
tan$=--~=OR
(v tan O" = 0)
~XL = Xc (Resonance) (1)
CHAPTER 7 : Alternating current
41. Figure shows the variation of t; with 00in a L-C-Rseries circuit for two values of resistance RI andR2(RI > R2),
R1.y R2'/ \'vV
I ~:..-f--V Wo
(j) (rad/s) - (1)
The condition for resonance in the L-C-R circuit is,I I
XL = Xc ~woL = -- ~wu =--woC LC
We see that the current amplitude is maximum atthe resonant frequency. Since im = Vm I R atresonance, the current amplitude for Ci1~C R2 issharper to that fOl '-,l~e RI•
Quality factor or simply the li-i,lllOl (If u n-sonantL-C-R circuit is defined as the ratio of voltage dropacross the resistance at resonance.Q=VL=WL
VR R
Thus finally, Q = .!...J~RvC
The" Q-factor determines the sharpnc-,s atresonance as for higher value of Q-fa(tof thetuning of the circuit and it sensitivity to acceptresonating frequency signals will be much highct ,
(2)
42. (i) The average power dissipated,P = (i2l?) = (i~Rsin2wt)= i~R(sin2(()t)
sin2wt =!. (l - cos 2(01)2
:. (sin2wt) =!. [l - (cos 2wt)] = ~2 2
(.: cos 2wt = 0)
43.
(ii) Power of the bulb, P = 100W and voltage,V = 220 V
The resistance of the bulb is given asR = V
2= p20)2 = 484 n
P 100 (1)
(i) When a source of AC is connected to acapacitor of capacitance C1 the charge on itglows from zero to maximum steady value Qo.
(2)
223The energy stored in a capacitor is
1 2E = -CVo
2where, Vo is maximum potential differenceacross the plates of the capacitor.The alternating voltage applied is
E = Eosinmtand the current leads the emf by a phase angleof 1t/2
1= loSin( wt+ ~)= Iocoswt
:. Work done over a complete cycle isT T
W = JEldt = J (Eo sinwt) (10 coswt) dto (J
E I T= _.lL...Q.J 2sinwlcosWidl2 0
W = ~2!!! r1sin2(,Hdt2 Jo
(1'12)
44.
(Ii) When DC ,,)UITe is connected, the condenseris charged but no current flows in the circuit.Therefore, the lamp does not glow. No changeoccurs even when capacitance of capacitor isreduced.When AC source is connected. the capacitoroffers capacitive reactance Xc = .2.... The
wecurrent flows in the circuit and the lamp glows.On reducing C1, x, increases. Therefore, theglowing of the bulb reduces. (1'12)
Given, L = 10 H, C = 40 j.lF,
R = 60 n. Vrms = 240 V(i) Resonating angular frequency,
1 1000 = -- = --p=====""".JLC ~IO x 40 x 10-6
00 - 1 50 rad/so - 20 x 10 ) (1)
(ii) Current at resonating frequency,
I = Vrms = Vrms (':At resonance, Z = R)rms Z R
= 240 = 4A60 (1)
(iii) .: Inductive reactance, XL = wL
At resonance, XL = woL =50 x 10 = soonPotential drop across to inductor,Vrms = Irills x XL = 4 x 500 = 2000 V (1)
224
45. Assuming XL> x, ~ VL> Vc
Net voltage, V = ~vi+ (VL - Vcl2
where, VL, Vc and VR are potential difference acrossL, C and R respectively.I+- VR=R1 -+t- VL=XL1 -+- Vc=XcI--I
II
c
V=Vo sin rot11
rn
But, VR = II?, VL = lXL '
Vc = tx,
V = ~(IR)2 + (!XL - !Xcl2
.!::: = ~R2 + (XL - Xcl2I
Impedance of L-C-R circuit,V I 2 2Z = - = vR + (XL - XclIVo
10=R
lI
46. Given, Vrms = 220 V, L = 20 mH = 2 x 10-2 H,
R =110 n.C = 800 ~F = 800 x 10-6 F
1t2 1t2
o Chapterwise CBSE Solved Papers PHYSICS
(i) Average power observed by L-C-R series ACcircuit is maximum when circuit is inresonance.. . Resonant frequency,
I000 = -JLC
IVo = 21t-JLC
21t 12X 10-2 x 800 X 10-6
V 1t2_ 1000 _ 125 -1
Vo ---- sz x 4
Vo = 125 S-1
(ii) As 1 = Vrms = 220 = 2A'rms Z 110
Z=R=110n ~ Irms=2A:. Maximum current amplitude,
10 = Irms ..[2 = 2..[2 A (1'12)
(2)
47. Due to change in flux, the emf is induced in the coil.The rate of change of flux will give the value of emf.
Let an alternating voltage, V = Vo sin 001 is appliedacross pure inductor of inductance L. Themagnitude of induced emf is given by
e = L dIdt (1)
(1)
For the circuit,Magnitude of induced emf = Applied voltage
. L dI H • dI Vo· t dt.e. -=Yosmoot or =-smoo Idt L
On integrating both sides, we get
1- Vo f' tdt- Vo (-cosoot)-- smoo -- ---L L 00
I = - Vo cos oot= - Vo sin (..:: - oot)ooL ooL 2
I = Vo sin (001 - ..::)XL 2
where, XL = ooL= inducting reactance
I = 10 sin ( oot- %) ... (ii)
(1)
or
... (i)
where, 10 = peak value of AC
But, V = Vo sin eor ... (iii)From Eqs. (ii) and (iii). it is clear that AC lagsbehind the voltage by phase ..::.
2 (1)
CHAPTER 7 : Alternating Current
48. (i) Let alternating voltage, V = Vo sin rot is appliedacross a capacitor C. At any instant, thepotential difference across the capacitor isequal to applied voltage. (1)
r------i r:C__ --,
L- ~.~,~----~
V= Vo sin rot ... (i):. V = Potential difference across the capacitor
=iC
~ q = CV or q = CVo sin (j)t
dq C I Vo.. - = (j) Vo cos (j)t or = -( ) cos (j)t~ 1
(j)C
[ = [0 sin ( (j)t + ~)
10 =~= Vo
(~) Xc
C . . Iapacitrve reactance, Xc =--roC (1)
(ii) From Eqs. (i) and (ii). current leads thevoltage by phase 2:-,
2
or ... (ii)
where,
V, Iv
Vo10
o r-----:,...;:..::.:=t---t--::-J---(j)-t
y
10
Vo
~------~~~-----x
49. (i) Since, resonant angular frequency (j)o is samefor both the graphs.
i.e.
225
(ii) (a)
or ~ = C2
L2 C1
(1)
is greater than
(b)
In graph (I), current t.;graph (II) 1m2•
i.e. 1m, > Im2 ~ Rl < R2Power factor of circuit represented by graph(I) has power factor as cos $ oc R. (1)
Q = ~ J¥ [For two graphs]
1Qoc_
Ri.e.
As, RJ < R2 :. q > Q2
Quality factor of graph (I) is higher thanthat of graph (II). (1)
50. (i) Device X is a capacitor.
As, the current is leading voltage by 2: radians.2
(1)
(ii) Curve A represents power,Curve B represents voltage andCurve C represents current.As, E (t)= Eo sinrotCurrent, I(t) =/0 cosrotAs, in the case of capacitor,
I = 10 sin ( (j)t+ ~) [current is leading voltage]
Average power, P= E (t) I (t) = E 010 cos $1 2 (2)
where, $= phase difference
(iii) As, Xc = capacitive reactance = ~Cro
where, (j) is angular frequency.So, reactance or impedance decreases withincrease in frequency.Graph of Xc versus (j) is shown below,Phasor diagram
E ----------. Eo
(1)
X' XCurrent leads emf by rc/2 radians
(iv) For a capacitor fed with an AC supply
V =ior q=CVC
I dq Eo . ( 1t)= dt = Xc sm (j)t+"2
226
51. If 1 is the current in the circuit containinginductor of inductance L, capacitor ofcapacitance C and resistor of resistance R in series,then the voltage drop across the inductor is
VL =/xXr!+----VR=Rl .t~ VL=XLI • I ~ Vc=Xcl~
R L
L---------------~~,r_------------~EoSincot
which leads current 1 by phase angle of 1t/2, andvoltage drop across the capacitor is V2 = 1 x Xc' (1)which lags behind current 1 by phase angle of 1t/2.and voltage drop across the resistor isVR = IR which is in phase with current I. So the
net voltage E across the circuit is (using phasordiagram)
VL
[n
Vc I VRPhasor diagram
E = ~vi+ (VL - Vel2
~ E =/~R2+(XL -Xel2 ~E = IZ
where, Z = ~R2 + (XL - Xel2 is known asimpedance. Phase angle between voltage and
v-v. X-Xcurrent is given by tane = _L C = _L __ C.
VR . R
A series L-C-R circuit has its natural angularfrequency, ro = ~
'\ILC
and natural (resonating) frequency, v = I~21t'\lLC
when the applied AC in the circuit has thisfrequency the series L-C-R circuit offers minimumimpedance i.e. only R and current at this frequencyflows maximum. In the case of resonance, voltageand current are in same phase.Above mentioned condition is known as conditionof resonance. In this condition (1)
(i) Inductive and capacitive reactances are equal
o Chopterwise CSSE Solved Papers PHYSICS
1 1XL=X~ ~roL=-~ro=--- [':ro=21tvl
" roC JLC1
~ V=---21tJLC
(ii) Potential drop across inductor and capacitorare equal, VL = Vc
(Hi) The series resonant circuit is also called anacceptor circuit because when a number ofdifferent frequency currents are into thecircuit, the circuit offers minimum impedanceto natural frequency current.For L-R circuit, XL = RPower factor, l'! = cose
R R 1
~R2 + X~ = R2 + R2 = .J2For L-C-R circuit, as C is put in series with L-Rcircuit and XL = Xc (3)
Power factor, P2 = cosoR R =~=1
~R2 + (XL - Xel2 ~R2 + (XL - XL)2 R
R . d . PI 1equue ratio = - = --P2 .J2
52. (i) To draw maximum current from a series L-C-Rcircuit, the circuit at particular frequencyXL = x.: (1)
v = __1_ = 1 = 39.80 Hz21tJLC zx 3l4.J8 x 2xI0-6
This frequency is known as the seriesresonance frequency.
(Ii) 1 - Eo _ 200 - 2A11 0------R 100 (1)
(ill)
Land CfixedR1 > R2
Frequency (1)
(iv) Sharpness of resonance It is defined as theratio of the voltage developed across theinductance (L) or capacitance (C) at resonanceto the voltage developed across theresistance (R).
Q=i~(1)
CHAPTER 7 : Alternating Current
It may also be defined as the ratio ofresonance angular frequency to thebandwidth of the circuit
Q=~Moo
Circuit become more selective if theresonance is more sharp, maximum currentis more, the circuit is close to resonance forsmaller range of (2~00) of frequencies. Thus,the tuning of the circuit will be good. (1)
53. (i) Refer to Ans. 51. (2)
(ii) L = -±-H, v = 50 Hz, R = lOOn, v = 200Y1t2
1',: XL =Xc orooL =-
ooc
C=_l_= 1002L 41t\.2 x L
1-------2500x16 40000
= 2. 5x10-5p = 251lP
I = .!::: = .!::: = 200 = 2AZ R 100
54. (i)
Vc- VL
Prom the phasor diagramV = VL VR + Vc
Magnitude of net voltage,
Vm = ~(VIm)2 + (Vcm - V1m)2
Vm = Im~[R?n + (Xc - XL)2]
Prom the figure,
tan <I>= v"m - ~m = 1m (Xc - XL)Vim 1mr
<1>=tan " (Xc ;XL)
(ii) Refer to Ans. 17.
227
(3)
55. (i) V = Vo sinoot and I = 10 sin ootWork done in small dt will be (2)
dW = P dt = VI dt = Volo sin200t dt~I= i..Q. (l - cos 2 oot) dt
2The average power dissipated per cycle in theresistor will be (2)
WITPay =T=yfdW
o~ I T
= i..Q. f (l - cos 2 oot)If 0
= Volo [t- Sin"lIDt]T- --- _.- J2T 200 0
= VOIO[(T_O)_O] .. 11, ,!,{,:Jl "JidIf ,;.J '-'1I1'J/; qoib 'J.);Jic. i
2 ~ li!.'O ".1 . t'l' '= Volo = Vo
2 2R_ ~oIo _ _ 1I:rm
2s r ~ ]P 11: I L': -20 = 1I:rmsav - .J2.J2 - rms rms - R(ii) The power is P = v'mslrms cos <1>.If cos <I>is small,
then current considerable increases whenvoltage is constant. Power loss, is 12 R. Hence,power loss increases. (112)
(iii) Refer to Ans. 9. (112)
56. Refer to Ans. 19 and for resonance condition referto Ans. 51. (5)
57. On increasing the frequency of AC. there will be no Ieffect on resistance (as it does not depend on i
frequencies). Inductive reactance will increase (as IXL =.21tfL or XL ec fl, capacitiv1 reactance will I
decrease (as Xc = _1_ or Xc oc -). If reSistance~r21tfC f
reactance of a circuit increases/decreases, thecurrent for the circuit decreases/increases .
._-_ ..._------------------
(1)
(i)
(2) I10
(1)
(1)
(j) ro, (j) - • "Jj (1)
The sharpness of resonance in series L-'C-R 'Cil'cuitrefers how quick fall of alternating current incircuit takes place when frequency of alternating
228
(t
voltage shifts away from resonant frequency. It ismeasured by quality factor (Q-factor) of circuit.The Q-factor of series resonant circuit is defined asthe ratio of the voltage developed across thecapacitance or inductance at resonance to theimpressed voltage which is the voltage applied.. Q all f Q voltage across L or Ci.e, u ty actor, = ------'=--------
applied voltage
Q = (wr L)lRl
[ .: applied voltage = voltage across R)
Q= wrLR
Q = (lk.orG)1 = _1 _Rl see;
Q= LRC·_1_
JLC
=i~
or Q=lJLC =.!.. fIRC RVC
Thus, Q=i~
This is the required expression.(ii) Let initially Ir current is flowing in all the
three circuits. If frequency of applied ACsource is increased, then the change in currentwill occur in the following manner.(a) Circuit containing resistance R
only There will not be any effect in thecurrent on changing the frequency of ACsource. (1)
or. (1)
[. 1usmg eo =--)
r JLC
Current I 1~r-------------
f; __ ----,__----:--
where, fi = initial frequency of AC source.There is no effect on current with theincrease in frequency.
(b) AC circuit containing inductance onlyWith the increase of frequency of ACsource inductive reactance increase as
I = Vrms = VnnsXL 21tfL
o Chapterwise CSSE Solved Papers PHYSICS
For given circuit,Ioe.!..
fCurrent II
I,
f;----f- (1)
Current decreases with the increase offrequency.
(c) AC circuit containing capacitor only
1 1Xc=-=--
c.oC 21tfC
Current, I = ;:s = (V71S )
21tfC
I = 21tfCVrms
For given circuit, I cc fCurrent increases with the increase offrequency.
f- (1)
58. Phase difference between voltage and current,X -Xtan o > L c ... (i)
Rand Io = Vo = Vo
Z ~(XL - Xcl2 + R2
:. Expression of AC,I = Io sin (wt - <p)
Conditions for resonance(1)
(i) Inductive reactance must be equal tocapacitive reactancei.e. XL = Xc
XL = Xc1
woL=-woC
(ii) As,
CHAPTER 7 : Alternating Current
2 1 1(00 = LC ~ (00 = .JLC
where, (00 = resonant angular frequency.(iii) Impedance becomes minimum and equal to
ohmic resistancei.e. Z = Zminimum = R
(iv) AC becomes maximum,
. . J = Vmax = Vmax
max Zmin R
(v) Voltage and current arrives in same phase.Power factori.e. cos <p= Pay
Vrms i-;True power
Apparent power
59.
The power factor is maximum.i.e. cos <p=+ 1, in L-C-R series AC circuit whencircuit is in resonance. The power factor isminimum when phase angle between V and Iis 90°, i.e. either pure inductive circuit or purecapacitive AC circuit. (1J
To find the voltage across each circuit element, isteps to be followed are: I
I
(i) To calculate the maximum current in circuit, Ifirstly find the (Z) impedance and rms value of icurrent. I
(ii) It can be calculated with the help of formulaof phase difference.
As, applied voltage
V = 140 sin 100 1t1
C= 50~,1t
C = 2 H, R = 400 n = 50 X 10-6 F1t 4
Comparing it with V = Vo sin (01,Vo = 140V, (0 = 100 1t (1J
Inductive reactance, XL = (OL5
XL = 1001t X - = 500 n1t
C. . 1apacinve reactance, Xc = -
weXc = = 200n
1001t x 50 X 104i1t
Impedance of the AC circuit, (1J
Z = ~R2 + (XL - Xcl2
= ~(400)2 + (500 - 200)2
229
(1J
Z = ~1600+ 900 = 500 n
Maximum current in the circuit,Io=Vo=140
Z 500I = Is. = 140 = 0.2 A
rms .fi 500 x Ji (1J
(1J
(1J
Vrms across resistor R, VR = Irms R
= 0.2 x 400 = 80 VVrms across inductor, VL = Irms XL
= 0.2 x 500 = 100 VVrms across capacitor, Vc = Irms x XC'
= 0.2 x 200 =)40 V
Here, V= VR + VL + VcBecause VL and VR are not in same phase,
.. V = ~vi+ (VL - Vcl2
60. (i) Refer to Ans. 31 (i). (3J
(ii) For definition of Q- factor refer to Ans.I.Q-factor should have high value in receiving. . Energy storedcircuits because Q = 21tJ x --:-''-'--;--
Power lossSo, higher the value of Q means the energy loss isat lower rate relative to the energy stored i.e. theoscillations will die slowly and damping would beless. Q-factor depends on(a) voltage across L (or C)(b) voltage across R
Q-factor =.!.. fI or ~R V-c (01- (02 (1J
Quality factor depends on ohmic resistance ofL-C-R series AC and inductance and capacitanceof L-C-R series AC circuit. (1J
(1J
(1J
61. For impedance Refer to Ans. 19.Also, from phasor diagram
tan <p= XL - Xc = Reactive impedanceR Resistance
'" -I [Reactive impedance]'I' = tan L Resistance
(1J
(1J
Impedance, Z = ~R2 + (XL - Xcl2
where, Z = R~ R = ~rR-=-2-+-(X-L-_-X-c-)72~ (XL - Xcl2 = 0
XL =XC1 1
~ (OoL=--~(Oo=--(OoC .JLC (3J
where, (00 is resonant L-C-R series AC circuit.
[TOPIC 2] AC Devices2.1 Chock CoilA chock coil is an electrical device which is usedfor controlling current in AC circuits withoutwasting electrical energy.Types of chock coil
(i) To reduce low frequency alternatingcurrents, chock coils with laminated softiron cores are used. These are called ofchock coils.
(U) To reduce high frequency alternatingcurrents, chock coils with air cores are used.TI1l'~C arc railed rfchock coils.
2 2 TransformerIt is a device which converts high voltage AC intolow voltage AC and vice-versa. It is based upon theprinciple of mutual induction. When a variablecurrent is passed through one of the twoinductively coupled coils, an induced emf is set upin other coil.
WorkingWhen an alternating current is passed throughthe primary coil, the magnetic flux through theiron core changes, which does two things,produces emf in the primary coil and an inducedemf is set up in the secondary coil. I f we as uinethat the resistance of primary coil is negligible,
then the back emf will be equal to the voltageapplied to the primary coil.
d<l> d<l>:. VI = -NI- and V2 = -N2-
dt dtwhere, NI and N 2 are number of turns in theprimary and the secondary coil respectively,while V] and V2 are their voltages, respectively.
Output emf = V2 = N 2
Input emf VI N I
Energy Losses in a Transformer(i) Eddy Current Loss Eddy current in iron
core of transformer facilitate the loss ofenergy in the form of heat.
(ii) Flux Leakage Total fluxes linked withprimary do not completely pass through thesecondary which denotes the loss in the fluxor flux leakage.
(iii) Copper Loss Due to heating, energy losstakes place in copper wires of primary andsecondary coils.
(iv) Hysteresis Loss The energy loss takesplace in magnetising and demagnetising theiron core over every cycle.
(v) Humming Loss The magnetostrictioneffect leads to set core in vibration which inturn produced the sound. This loss isreferred as humming loss.
Types of TransformerThere are two types of transformers which aregiven as below:
(i) Step-up transformer (NI > N2) Itconverts low alternating voltage into highalternating voltage.
(ii) Step-down transformer (NI < N2)
it converts high alternating voltage into lowalternating voltage.
Important Points Related toTransformero For an ideal transformer,
Input power = Output power
VIII = V2I2 ~ VI = 2v2 II
o Transformation Ratio (r)
r=Ns=Vs=[pNp Vp Is
where, N pand N s are the number of turns inprimary and secondary coils. Vp and Vs arealternative voltage in primary and secondarycoils.I p and Is are AC in primary and secondarycoils.
o Long distance power transmission takesplace at high alternating voltage so as tominimise losses in the form of heat,therefore, step-up transformers are used atpower stations and step-down transformersat receiving ends.
o Efficiency of transformer,11 = Output power x 100
Input powero The flux linked with the coil at any instant t
is given by Q> == NBA cas O){
~ Q>max = NBA [.: (cos O)t)mdx = 1]o Instantaneous value of alternating voltage is
given by V == N BA (I)
o Peak value of alternating voltage is given byVo = NBAO)
o Instantaneous AC is given by [= [0 sin O)t
[0-_ NBAO).where,
R
where, N = number of turns, B = magneticfield, A = area and 0) = frequency
PREVIOUS YEARS'EXAMINATION QUESTIONSTOPIC 2
o 1 Mark Questions1. Why is choke coil needed in the use of
fluorescent tubes with AC mains? Dalhl2014
2. Why is the core of a transformer laminated?Delhl2013C
3. Mention two characteristic properties of thematerial suitable for making core of atransformer. All Indio 2012
4. What is the function of a step-uptransformer? All Indio 2011C
5. Write any two factors responsible for energylosses in actual transformers. Delhi 2009C
6. A power transmission line feeds input powerat 2200V to a step-down transformer with itsprimary windings having 3000 'turns. Find thenumber of turns in the secondary winding toget the power output at 220 V. Oelhl2017
o 2 Marks Questions7. State the underlying principle of a transformer.
How i he large scale transmission of electricenergy over long distances done with the use oftransformers? All Indio 2012
8. A power transmission line feeds power at2200 V with a current of 5 A to step downtransformer with its primary winding having4000 turns. Calculate the number of turnsand the current in the secondary in order toget output power at 220 V.Foreign 2011
232
9. State the principle of working of atransformer. Can a transformer be used tostep-up or step-down a DC voltage?Justify your answer. All India 2011
10. Mention various energy losses in atransformer. All India 2011
o 5 Marks Questions11. A horizontal conducting rod 10 m long
extending from East to West is fallingwith a speed 5.0 ms-1 at right angles tothe horizontal component of the earth'smagnetic field, 0.3 x 10-4 Wb m -2. Findthe instantaneous value of the emfinduced in the rod. All India 2017
12. (i) Draw a labelled diagram of astep-down transformer. State theprinciple of its working.
(ii) Express the turn ratio in terms ofvoltages.
(iii) Find the ratio of primary andsecondary currents in terms of turnratio in an ideal transformer.
(iv) How much current is drawn by theprimary of a transformer connected to220 V supply when it delivers powerto a 110 V-550 W refrigerator?All India 2016
13. (i) Write the function of a transformer.State its principle of working withthe help of a diagram. Mentionvarious energy losses in this device.
(ii) The primary coil of an ideal step uptransformer has 100 turns andtransformation ratio is also 100. Theinput voltage and power arerespectively 220 V and 1100 W.Calculate(a) number of turns in secondary(b) current in primary(c) voltage across secondary(d) current in secondary(e) power in secondary Delhi 2016
14. (i) Draw a schematic arrangement forwinding of primary and secondarycoils in a transformer when the twocoils are wound on top of each other.
o Chnpterwise eBSE Solved Papers PHYSICS
(ii) State the underlying principle of atransformer and obtain theexpression for the ratio of secondaryto primary voltage in terms of the(a) number of secondary and primary
windings and(b) primary and secondary currents.
(iii) Write the main assumption involvedin deriving the above relations.
(iv) Write any two reasons due to whichenergy losses may occur in actualtransformers. All India 2014C
15. (i) State the principle of a step-uptransformer. Explain with the help ofa labelled diagram, its working. .
(ii) Describe briefly and two energy lossesgiving the reasons for their occurrencein actual transformer. Foreign 2012
16. (i) With the help of a labelled diagramdescribe briefly the underlying ,principle and working of a step-uptransformer.
(ii) Write any two sources of energy lossin a transformer.
(iii) A step-up transformer converts a lowinput voltage into a high outputvoltage. Does it violate law ofconservation of energy? Explain.Delhi 2011
17. (i) Draw a schematic diagram of astep-up transformer. Explain itsworking principle. Assuming thetransformer to be 100% efficient.obtain the relation for(a) the current in the secondary in
terms of the current in theprimary and
(b) the number of turns in theprimary and secondary windings.
(ii) ~ention two important energy losses10 actual transformers and state howthese can be minimized?All India 2009C; Foreign 2009
18. Draw a schematic diagram of a step-uptransformer. Explain its workingprinciple. Deduce the expression for thesecondary to primary voltage in terms ofthe number of turns in the two coils.In an ideal transformer, how is this relationrelated to the currents in the two coils.
CHAPTER 7 : Alternating Current
How is the transformer used in large scaletransmission and distribution of electricalenergy over the long distances? All India 2010
19. A step-up down transformer operated on a2.5 kV line. It supplies a load with 20 A.The ratio of the primary winding to thesecondary is 10 :1. If the transformer is90% efficient, calculate
(i) the power output(ii) the voltage and
(iii) the current in the secondary coil.Foreign 2010
o Explanations1. A choke coil is needed in the use of fluorescent
tubes with AC mains because it reduces thevoltage across the tube without wasting muchpower. It is an inductor with large inductance toreduce current in AC circuits without much lossof energy. (1)
2. The core of transformer is laminated to reduce theenergy losses due to eddy currents, so that itsefficiency may remain near 100%. (1)
3. The characteristic properties of the materialsuitable for making core of a transformer are asfollow:(i) Low retentivity or coercivity.
(ii) Low hysteresis loss or high permeability andsusceptibility. (1)
4. Step-up transformer converts low alternatingvoltage into high alternating voltage and highalternating current into low alternating current. Thesecondary coil of step-up transformer has a greaternumber of turns than the primary coil(Ns > Npl. (1)
5. Two factors for energy losses are(i) eddy currents (112)
(ii) hysteresis loss (1/2)
6. Given, input power (Vp) = 22.00 VNumber of turns (n1) = 3000Output power (Vs) = 220 VAs, V, = n2 => 220 = ~
Vp n1 2200 3000
220=> n2 = -- x 30002200
. . Number of turns in the secondary winding,n2 = 300 turns. (1)
233
7. Principle of transformer A transformer isbased on the principle of mutual induction,i.e. whenever the amount of magnetic flux linkedwith a coil changes, an emf is induced in theneighbouring coil. This changing flux sets up aninduced emf in the secondary coil, also selfinduced emf in primary coil. (1)Power transmission Electric power istransmitted over long distances at high voltage. So,step-up transformers are used at power stations toincrease the voltage of power whereas a series ofstep-down transformers are used to decrease thevoltage upto 220 V. (1)
8. Vp = 2200V, t, = 5A, n, = 4000
Vs =220V,Ns =?Is =?Vs = Ip = NsVp Is n, (1)
220 = 2. = l!.L2200 Is 4000
220 5 1 5-- = - => - = - => Is = 50 A2200 Is 10 Is
5 Ns 5 Ns-=-- => -=-- => Ns =400Is 4000 50 4000 (1)
9. (i) For principle of working of transformerRefer to Ans. 7. (1)
(ii) No, transformer cannot be used to change DCvoltage because DC voltage cannot change fluxlinked with primary or secondary coils. (1)
10. For energy losses in transformer. (1)Refer to Ans. 5 (1)
11. Instantaneous emf induced in conducting rode = Blvvolt
(As rod is falling perpendicularly to B)Here, B = BH = 0.3 X 10-4 T
I = 10mv= 5ms-1
(1)
So, induced emf,e = o.3xl0-4 x10 x 5= 1.5 X 10-3 volt = 1.5mV (1)
12. (i) Step down transformer:Core
(2)
234
Principle: When the current flowing throughthe primary coil changes, an emf is induced inthe secondary coil due to the change inmagnetic flux linked with it i.e., it works onthe principle of mutual induction.For step down transformer,Ns < Np, hence Es < lOp'
(ii) !2. = NslOp Np
(iii) For an ideal transformer, Pin = POUl
lIONEpIp = e.t, ~ J!...=-,-=_sIs lOp Np
13.
(iv) Pjn = Pout = 550W ~EpIp = 550
220xI =550~I =550=~=2.5Ap p 220 2 [1)
(i) Refer to text Page 225-226There are number of energy losses in atransformer. [1)
(a) Copper losses due to Joule's heatingproduced across the resistances of primaryand secondary coils. It can be reduced byusing copper wires. [1)
(b) Hysteresis losses due to repeatedmagnetisation and demagnetization of thecore of transformer. It is minimised byusing soft iron core, as area of hysteresisloop for soft iron is small and hence energyloss also becomes small. [1)
(c) Iron losses due to eddy currents producedin soft iron core. It is minimised by usinglaminated iron core.
(d) Flux losses due to flux leakage orincomplete flux linkage and can beminimised by proper coupling of primaryand secondary coils.
(ii) Here, Np = 100, Ns = 100Np
Ej = lOp = 220 V, PI = 1100 W(a) Np = 100:.Ns = 10000(b) i, = PI = 1100 = 5A
lOp 220N(c) Es = _s X lOp = 100 x 220 = 22000 Vn, [2)
(d) Is = Po = 1100 = 2.. A (.: Po = PI)Es 22000 20
(e) Ps = Po = PI = 1100 W
o Chapterwise CBSE Solved Papers PHYSICS
14. (i) The schematic arrangement of a transformer isshown as below: [1)
e-roEit
z-ro"0cooQ)
(f)
[1)
[1)
(ii) Principle of transformerRefer to Ans. 6. (2)
working When an alternating current ispassed through the primary, the magnetic fluxthrough the iron core changes, which does .two things, produces emf in the primary andan induced emf is set up in the secondary. Ifwe assume that the resistance of primary isnegligible, then the back emf will be equal tothe voltage applied to the primary.
(a) .. 1-) = - N) dcp and V2 = - N/CPdt dt
where, NI and N 2 are number of turns inthe primary and the secondary coilsrespectively while 1-) and V2 are theirvoltages respectively.
(b) But for ideal transformers,VI II = V2I2 ~ 1-) = 12
V2 II
(iii) Main assumptions(a) The primary resistance and current are small.(b) The flux linked with primary and
secondary coil is somewhere is no leakageof flux from the core.
(c) secondary current is small.(iv) Refer to Ans. 13. (2)
15. (i) Principle of transformerRefer to Ans. 5. [2)
(3)
CHAPTER 7 : Alternating Current
16. (i) Refer to Ans. 14. (3)
(ii) Refer to Ans. 14 (ill). (1)
(ill) No, it does not violate the law of conservationof energy because voltage increase isaccompanied by decrease in current.In step-up transformer the current decreasesby the same proportion as the voltageincreases. When voltage increases 11 times, thecurrent reduces to I/n time. (1)
17. (i) Refer to Ans. 14. (3)
(a) Let alternating current, alternating voltageand number of turns in primary andsecondary coils are (IpJs), (Vp, Vs) and(Np,Ns)' respectively.r , 100% efficient transformer,Input power = Output power~ VpIp = Vs IsAlso, assuming transformation ratio oftransformer is r.
r Ns=Vs=IpN p Vp Is
~ i, VSXIsVp (112)
(b) r = NsNp (112)
(ii) Two energy lu~"~s in the actual transformers(a) Hysteresis loss Loss of energy in
magnetising and demagnetising the C( • oftransformer in every cycle. It can beminimised by taking core of soft iron 1 inglow coerdvity, retentivity and hysteresis loop.
235
(b) Eddy current Energy loss in the form ofheat due to eddy current. It can beminimised by taking laminated coreconsisting of insulated rectangular sheets,piled-up one over another. (1)
18. Refer to Ans. 14 and 17.Step-up transformers are used at generatingstations so as to transmit (he power at highvoltage to minimise the loss in the form of heat,whereas series of step-up trransformers are usedat receiving ends. (2)
19. Given, input voltages, v, = 2.5 X103V (3)
Input current, Ip = 20AAlso, Np =10 ~ Ns =~
Ns I n, 10Percentage efficiency = Output power x 100
Input power~ = Output power100 VpIp
(i) Output power = ~ x (Vp I p)100
=~X(25XI03)X(20A) =4.5XI04 W100
(ii) :. Vs = Ns ~ Vs = Ns x v,Vp n, Np
Voltage, Vs =~X2.5XI03V=250V10
(iii) VsIs =4.5xI04W
4.5X104Current, Is = ---
Vs
(3)
... (i)(1/2)
(112)
(1/2)
4.5x104
~ Is =J80A250 n/~
Value Based Questions [From Complete Chapter]o 4 Marks Questions
1. Sushil is in the habit of charging his mobileand then leaving the charger connectedthrough the mains with the switch on.When his sister Asha pointed it out him, hereplied there was no harm as the mobilehad been disconnected. Asha thenexplained to him and convinced him, howthe energy was still being wasted as thecharger was continuously consumingenergy. Answer the following questions.
(i) What values did Asha display inconvincing her brother?
(ii) What measures in your view, should beadopted to minimise the wastage ofelectric energy in your households?
(iii) Imagine an electric appliance of 2 W,left connected to the mains for20 hours. Estimate the amount ofelectrical energy wasted. Foreign 2015
Ans.(i) Asha displayed the values of awareness towardsenergy saving and her concern towards wastageof energy. (1)
(ii) Following measures should be adopted tominimise the wastage of electric energy in ourhouseholds:(a) We should switch off the electrical appliances
which are not in use.(b) The old electrical appliances should be
repaired time to time, as they often consumemore electricity. (1)
(c) Condensers of electrical appliances should bereplaced time to time.
(d) New electrical and electronic appliances mustbe purchased in accordance with theirefficiency to consume less energy.
(e) Use of solar devices like solar heater andcookers must be encouraged rather than theuse of conventional electrical appliances. (1)
(iii) Electrical energy dissipated by an appliance isgiven by E = P x twhere, P is the power and t is the time.
P=2W,t=20h=20x60x60s (1)E = 2 x 20 x 60 x 60E = 1.44 x 105 J
So,
2. A group of students while coming fromthe school noticed a box marked "DangerHT 2200 V" at a substation in the mainstreet. They did not understand theutility of such a high voltage, while theyargued, the supply was only 220 V. Theyasked their teacher this question thenext day. The teacher thought it to be animportant question and thereforeexplained to the whole class.Answer the following questions:
(i) What device is used to bring thehigh voltage down to low voltage ofAC current and what is theprinciple of its working?
(ii) Is it possible to use this device forbringing down the high DC voltageto the low voltage? Explain.
(iii) Write the values displayed by thestudents and the teacher.All Indio 2015
Ans.(i) The device that is used to bring high voltagedown to low voltage of an AC current is atransformer. It works on the principle ofmutual induction of two windings orcircuits. When current in one circuitchanges, emf is induced in the neighbouringcircuit. (1)
(ii) The transformer cannot convert DC voltagesbecause it works on the principle of mutualinduction. When the current linked with theprimary coil changes, the magnetic fluxlinked with secondary coil also changes. Thischange in flux induces emf in the secondarycoil. If we apply a direct current to theprimary coil, then current will remainconstant. Thus, there is no mutual inductionand hence, no emf is induced. (2)
(iii) The value of gaining knowledge andcuriosity about learning new things is beingdisplayed by the students. The value ofproviding good education and undertakingthe doubts of students has been displayed byteacher. (1)
CHAPTER 7 : Alternating Current
3. One morning an old man walked bare-footto replace the fuse wire in kit-kat, fittedwith the power supply mains for hishouse. Suddenly he screamed andcollapsed on the floor. His wife criedloudly for help. His neighbour's son Anilheard the cries and rushed to the placewith shoes on. He took a wooden bat andused it to switch off the main supply.Answer the following questions
(i) What is the voltage and frequency ofmains supply in India?
(ii) These days most of the electricaldevices we use require AC voltage.Why?
(iii) Can a transformer be used to step upDC voltage?
(iv) Write two qualities displayed by Anilby his action. Delhi 2015
Ans.(i) In India, the voltage and frequency mainssupply is 220 V and 60 Hz respectively. (1)
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(ii) Since, the power loss at 220 V supply is lessthat at 110 V. So, due to this reason, most ofthe electrical devices require AC voltage. (1)
(ill) Since, the transformer requires an alternatingmagnetic flux to operate correctly,transformers cannot therefore be used totransform or supply DC voltages or currents.Also, the magnetic field must be changing toinduce a voltage in the secondary winding.Even if in case, the transformers primarywinding is connected to a DC supply, theinductive reactance of the winding. Would bezero as DC has no frequency. So, the effectiveimpedance of the winding will therefore bevery low and equal only to the resistance ofthe copper used. Thus, the winding will drawa very high current from the DC supplycausing it to overheat and eventually burnout,
because as, we know I = ~.R (1)
(iv) Anil is very helpful and has a great presenceof mind as in such condition, he used thewooden baton to switch off the main supply.
(1)
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