Circuit Theory - Solved Assignments - Semester Spring 2008

Solution of Assignment 1(Spring 2008)CIRCUIT THEORY (PHY301)

MARKS: 35

Question #1:We have usually two sources of electricity Main & Batteries. Here you are given someevery day home items categories these in such a way, so that we may understand whichone is operated from Main which from Batteries and those operated from both sources.I) Clock II) Kettle III) Light Bulb IV) Radio V) Dishwasher VI) Mobile Phone VII)Fridge.

Solution1. Clock can run on both mains electricity & batteries.2. The kettle uses mains electricity.3. Light bulb uses mains electricity.4. Radio can run on both mains electricity & batteries.5. Dishwashers use mains electricity.6. Mobile phone can run on batteries but charge up using mains electricity.7. Fridge uses mains electricity.

Question #2:Find V1 & V2 through the following circuit given below .Draw the circuit diagram of

each step otherwise you will lose your marks. Write each step of the calculation to getmaximum marks and also mention the units of each derived value

Question #3:Find total current I passing through circuit shown below. Draw the circuit diagram ofeach step otherwise you will lose your marks. Write each step of the calculation to getmaximum marks and also mention the units of each derived value

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Assignment 2(Spring 2008) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 50

Due Date: 05/05/2008

Q.1. You are given the circuit in Fig (A), Fig (B) & Fig (C).

a) You will find out node voltages V1,V2,V3 for each diagram. b) You will find out currents IA,IB,IC& ID for each diagram. c) You will also find out VA,VB,VC for each diagram.

From this study, state what you have learned about changing Reference Node in Nodal Analysis. Note: Draw and labeled complete circuit diagram otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. Part (A):

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Write KCL equations at V1.

1 2

2V V−

+ 1 3

4V V−

- 3 = 0

2 - 2 + - -12 = 0 1V 2V 1V 3V

3 - 2 - = 12 ________ (1) 1V 2V 3V Write KCL equations at V2.

2

2V V− 1 + 2 3

8V V−

+ 2

4V

= 0

-4 + 7 - = 0 ________ (2) 1V 2V 3V Write KCL equations at V3.

3 2

8V V−

+ 3 1

4V V−

+ 2 AI = 0

We know

AI = 1 2

2V V−

______ (A)

- +2 -2 + (23V 2V 3V 1V AI )8 = 0 _______ (B)

Put (A) in (B)

- +2 -2 + 8 -8 = 0 3V 2V 3V 1V 1V 2V 6 -9 +3 = 0 _______ (3) 1V 2V 3V Solving (1),(2) and (3) simultaneously

= 4.0V ; =2.4V and = -2.4V 1V 2V 3V

AI = 1 2

2V V−

= 1.2 A

BI = 1 3

4V V−

= 7.24

= 1.8A

CI = 2 3

8V V−

= 2.4 2.4

8+

= 0.6A

DI = 2

4V

= 0.6A

Now For VA =1 3V V− AV

= =4.8-(-2.4) AV 1 3V V− AV =7.2V



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For VB = =2.4-(-2.4) BV 2V V− 3

=4.8V BVFor VC = CV 1V =4.8V CV Part (B) Write KCL equations at V1.

1

2V

+ 1 3

4V V−

-3 = 0

2 + - =12 1V 1V 3V 3 - =12 ----------- (1) 1V 3V Write KCL equations at V2.

2

4V

+3-2 AI = 0

AI = 1

2V

2

4V

+3-2( 1

2V

) = 0

+12-4 = 0 2V 1V

-4 = -12 ------------- (2) 2V 1V Write KCL equations at V3.

3

8V

+ 3 1

4V V−

+2 AI = 0

AI = 1

2V

3

8V

+ 3 1

4V V−

+2( 1

2V

) = 0

V +2V -2V +8V = 0 3 3 1 1



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6 +3 = 0 1V 3V =2.4V ; =-2.4V ; =-4.8V 1V 2V 3V

AI = 1

2V

=1.2A

BI = 1 3

4V V−

=2.4 4.8

4+

=1.8A

CI =- 3

8V

=4.88

=0.6A

DI =- 2

4V

=2.44

=0.6A

For VA

AV = BI x 4= 1.8 x 4 = 7.2V

For VB

=8BV CI = 4.8V

For VC

- = 0 1 3V V− 2V =3V 1 2V V− = 2.4-(-2.4) =4.8V

Part (C) Write KCL equations at V1.

1 2

4V V−

+ 1 3

8V V−

+ 1

2V

= 0

2V -2V +V -V +4V = 0 1 2 1 3 1

7V -2V -V = 0 ------------- (1) 1 2 3 Write KCL equations at V2.



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2 1

4V V−

+ 3 -2 AI = 0

Where AI = - 1

2V

2 1

4V V−

+3-2(- 1

2V

) = 0

- +12+4 = 0 -------------- (2) 2V 1V 1V Write KCL equations at V3.

3 1

8V V−

+ 3

4V

+2 AI = 0

Where AI = - 1

2V

3 1

8V V−

+ 3

4V

+2(- 1

2V

) = 0

+23 1V V− 3 18V V− = 0

19 3V V3+ = 0 ------------ (3)

Solving eq (1), (2) and (3) simultaneously

= -2.4V , =-4.8V and =-7.2V 1V 2V 3V

AI =- 1

2V

=2.42

=1.2A

BI =- 3

4V

=-( 7.2)

4−

=1.8A

CI = 1 3

8V V−

=2.4 ( 7.2)

8− − −

=0.6A

DI = 1 2

4V V−

=2.4 4.8

4− +

=0.6A

For VA =AV BI x4=1.8x4=7.2V

For VB =BV CI x8=0.6x8=4.8V

For VC = - = -(-4.8)= 4.8V CV 2VOverall summary

Part (A) Part (B) Part (C)

1V 4.0V 2.4V -2.4V

2V 2.4V -2.4V -4.8V

3V -2.4V -4.0V -7.2V

AI 1.2A 1.2A 1.2A

BI 1.8A 1.8A 1.8A

CI 0.6A 0.6A 0.6A

DI 0.6A 0.6A 0.6A



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AV 7.2V 7.2V 7.2V

BV 4.8V 4.8V 4.8V

CV 4.8V 4.8V 4.8V

Conclusion: Changing Reference nodes, changes the reference node voltages. However, all currents and voltages across elements remain the same.

Q.2. Use nodal analysis to find Current Ib in the network given below. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. Write KCL equations at Vx.

24

250xV V−

+50

xV+

60150

x bV I− = 0

3 -72+15 -300xV xV bI =0 ---------- (I)

bI =24

250xV−

------------(A)

Put A in (I) So,

23 -300(xV 24250

xV−) = 72

23 - 28.8 +1.2 = 72 xV xV 24.2 = 100.8 xV = 4.165V xV



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bI = 24

250xV−

= 24 4.165

250−

=79.3mA

Q.3. First Identify and label each node in the network. Use nodal analysis to find voltage at each node in the network given below. Draw and labeled complete circuit diagram otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. First we will identify nodes and labeled them, Write KCL equations at V1.

1 122

V −+ 1 2

8V V−

+3+ 1

4V

= 0

4 -48+ - +24+2 = 0 1V 1V 2V 1V 4 - = 24 ----------- (A) 1V 2VWrite KCL equations at V2.



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2 1

8V V−

+ 2 51

V V0+-3 = 0

+8 +40 -24 = 0 -------------- (1) 2 1V V− 2V 0VBut

1V V0+ -12= 0

= 12- ---------- (2) 0V 1VPut (2) in (1)

- + 9 +40(12- )=24 1V 2V 1V -41 +9 = -450 ------------- (B) 1V 2VSolving (A) and (B)

= -10.91V , =-100.36V 1V 2V

------ Good Luck -----



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Due Date: 16/05/2008

Q.1. First Identify and label each node in the network. Use nodal analysis to find voltage at each node in the network given below. Draw and labeled complete circuit diagram otherwise you will lose your marks. Write each step of the calculation to get maximum marks and also mention the units of each derived value.

Sol. First we will identify nodes and labeled them,

Write KCL equations at the super node

1+ 2 = 0V 1

4V

+ 2

1V

+ 1 3

1V V− ----------------- (1)

But =0V 1 3V V−



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Hence eq. (1) becomes

4=-8 + 4 -4 +8 + +4 1V 2V 3V 3V 1V 1V 4=-3 + 4 +4 -------------------- (2) 1V 2V 3VWrite KCL equations at node V3.

+02V 3

4V

= 1 3V V− + 3102

V−

3

4V

+ 3 102

V −+ 3 1

1V V−

- =0 02V

4 - =20 --------------- (3) 1V 3VAt the super node constraint equation will be

V V ------------- (A) 2 1 4− = 0I

But 0I = 3

4V

Hence from constraint eq. we have

= +V -------------- (4) 2V 1V 3Solving eq.(4) and (2) we get

+8V =4 -------------- (5) 1V 3 Solving eq. (5) and eq. (3)

V = -0.12v 3Put this value in eq.(3) we get

=4.97v 1VAs

= +V 2V 1V 3Putting values we get

V =4.85v 2

Q.2. Use Mesh analysis to determine Currents in each Mesh in the network given below. Identify and label each mesh otherwise you will lose your marks. Draw and labeled complete circuit diagram. Write each step of the calculation to get maximum marks and also mention the units of each derived value.



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Sol. First we will identify mesh and labeled them,

For mesh I

-10-2 xI +10 1I -6 2I =0

But

xI = 1I - 2I

Hence,

10=-2 1I +2 2I +10 1I -6 2I

5=4 1I -2 2I ------------- (1)

For mesh II

12+8 2I -6 1I =0

6=3 1I -4 2I ----------- (2)

Solving (1) and (2)

1I =0.8A , 2I =-0.9A

Q.3. Use Mesh analysis to find Voltage Vx and Ix in the given network. Identify and label each mesh otherwise you will lose your marks. Draw and labeled complete circuit diagram. Write each step of the calculation to get maximum marks and also mention the units of each derived value. .



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Sol. Now redraw above equation with current sources removed. KVL for super mesh

-50+ 10 1I +5 3I +4 XI =0

Note that

xI = 1I

14 1I +0 2I +5 3I =50 ---------------- (1)

Constraint equation at X:-

2I - 1I =3

So, 1I - 2I +0 3I =-3 --------------- (2)

Constraint equation at Y:-

3I - 2I = 4xV

------------------- (A)

From the fig

=2(xV 1I - 3I )

Put in(A)

2I +2 1 3( )4

I I−- 3I =0

2I +0.5 1I -0.5 3I - 3I =0

0.5 1I + 2I -1.5 3I =0 --------------- (3)

Solving (2) and (3) we get

1I =-2+ 3I

Put this value in eq.(1),we get

3I =4.10A



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Put this value in eq.(1) we get

1I =2.10

By putting I1 in eq.(2)

2I =5.1A

VX = 2(I1-I3) = 2(2.11 -4.11) = -4V

------ Good Luck -----



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Due Date: 28/06/2008

Q.1. Find Vo by Superposition. Draw and label the circuit diagram of each step, otherwise you will lose your marks. Write each step of calculation to get maximum marks, also mention the units of each derived value. Solution: When only 12v source is acting: We replace 9v by short circuiting

Now here 3k and 3k are in parallel so 1.5K and 6k are in series so 1.5+6=7.5kΩ 3 3 1.5

3 3x

=+

ΚΩ

7.5k and 3k are in parallel 7.5x3/7.5+3=2.14ΚΩ V2.14=12x2.14/3+2.14= 5v



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Since 2.14k is equivalent of parallel 3k and 7.5, so same 5v voltage drops across 3k so Vo1= -5v (-ve sign is due to opposite polarity of Vo and 12v voltage source)

When 9v source is acting alone: We remove 12v and replace it by short circuit, circuit will be as

Method 1: 3k and 3k at right side are in parallel so 3x3/3+3=1.5ΚΩ

1.5k and 6k are in series, 1.5+6=7.5ΚΩ 7.5k and 3k are in parallel 7.5x3/7.5+3=2.14ΚΩ

By voltage divider rule, voltage across 2.14k is V2.14=2.14x9/3+2.14 V2.14=3.74v Since 2.14k is the equivalent of parallel 7.5k and 3k , same voltage drops across 7.5k And 7.5 is the equivalent of series 6k and 1.5k, so voltage drop across 1.5 will be V1.5=1.5x3.74/6+1.5=0.75v As 1.5k is equivalent of parallel 3k and 3k , so same 0.75v will drop across required 3k. Vo2=0.75v



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2nd method (by nodal analysis) Vo across 3k can be find out by Nodal analysis method By nodal analysis: Labeling nodes At node 1 Writing nodal equation at node 1 V1/3+(v1-9)/3+(v1-v2)/6=0 2v1+2v1-18+v1-v2=0 5v1-v2=18…………….(1) At node 2 Writing nodal equation at node 2 V2/3+v2/3+(v2-v1)/6=0 2v2+2v2+v2-v1=0 -v1+5v2=0…………….(2) Multiplying (2) by 5 and adding in (1) we get V2=0.75v As V2=Vo Vo2=0.75v By super position, Vo is due to combined effect of each 12v and 9v voltage source i.e. Vo=Vo1+vo2

V o= -5+0.75 Vo = -4.25v

Q.2 Find Io by Source transformation. Draw and label the circuit diagram of each step, otherwise you will lose your marks. Write each step of calculation to get maximum marks, also mention the units of each derived value.



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Rearranging the circuit in straight form

Using source transformation, 3mA into voltage source

.

4kΩ and 4kΩ are in series so Transform 12V , 8kΩ into 8kΩ in parallel with 1.5mA 8kΩ , 10kΩ are in parallel



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Converting parallel 4.44k and 1.5mA into voltage source, we get 6.67v so

Combining Voltage source, 12v- 6.67v=5.33v

Transforming 5.33v into current source

12kΩ , 4.44kΩ are in parallel

Current through 6k using Current division rule,

0.421oI mA= Answer



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3. Using superposition a) Determine the short-circuit current, that is, the current flowing through a wire

with zero ohm connected between a-b. b) Determine the open-circuit voltage, that is, the voltage between a-b with the short removed. Solution: We know superposition theorem is applicable to independent source and if circuit contains dependent source, we apply superposition theorem without removing or changing dependent source.Therfor in this circuit we calculate required values due to effect of 10A current source without removing dependent voltage source of 2Vx. (a) To find short circuit current Isc , first of all we assign node voltage. As Vx of 5Ω is directly attached to node1 so node voltage of it is Vx and same Vx drops across node2.so Writing node equations Vx/5+(Vx-2Vx)/20+Vx/10=10 4Vx+Vx-2Vx+2Vx=200 5Vx=200 Vx=40v As Isc=Vx/R=40/10 Isc =4A (b) Finding open circuit voltage



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Vx/5+(Vx-2Vx)/20=10 3Vx=200 Vx=200/3 Vx=66.7v As Voc=Vx so Voc=66.7v …………………….Good Luck………………….



Assignment 5(Spring 2008)(Solution)

CIRCUIT THEORY (PHY301)MARKS: 30

Due Date: 10/07/2008

Q.1.Use Thevenin’s theorem to find Vo in the network given below.Show each step of calculation otherwise you will lose your marks. Draw and label thecircuit diagram of each step and also mention the units of each derived value.

Solution:Finding Vth: To find Vth we remove RL , so circuit will be

2

Vth will be voltage drop across 2kΩ and 12v.As current flowing through 2k is 2mA so

V2k=2kx2mAV2k=4v

Now Vth=V2k+12vVth=4+12Vth= 16v

Calculating RthTo find Rth we short circuit voltage source and open circuit current

source

Adding series 1k and 1k and

Rearranging the circuit we have

2KΩ branch is in parallel with the short circuited branch. so we can neglect that 2KΩbranch therefore Rth=2kΩ

Thevenin’s theorem circuit for the given circuit is,

From this Voltage across 2KΩ Resistor is

Q.2.

Use Norton’s theorem to find Io in the network given below.Show each step of calculation otherwise you will lose your marks. Draw and label thecircuit diagram of each step and also mention the units of each derived value.

Sol:We have to find Io through 2kΩ so according to Norton theorem we follow these stepsFinding IN:

We short circuit the load resistance of 2k.

Converting the current source to voltage source

V2 = (3k)(2m) = 6VIN = I1 + I2

Finding Rth:To find RN short the voltage source and open the current source.

3k and 3k in series and in parallel to 6k

RN = 6k || (3k+3k)

The Norton's equivalent circuit will be

Using current divider rule

Q.3.How narrow and wider layer is produced in semiconductor .Do you think narrow orwider depletion region in semiconductor cause more barrier voltage, give reason.Answer: 5If the semiconductor is in forward bias and the applied voltage is greater than the barrier potentialthe electrons in the N region are repelled by negative terminal of the battery and moves towardsthe junction similarly the holes in P-region moves towards the junction .Majority charge carriersare cross the junction and the minority charge carriers are attracted by the battery. in this casedepletion layer is decrease. Then the barrier potential is nearly zero. Similarly the semiconductoris connected in reverse bias then the depletion is increases and barrier voltage is increases thebarrier voltage depends on the temperature of the junction, nature of semiconductor and dopingconcentration.

When semiconductor in forward bias, the depletion area is narrowed. the same butopposite can be said about reverse bias the depletion region is wider and the potentialbarrier is increased. At equilibrium potential barrier is V0 in FW barrier is V0-Vfw forRB potential barrier is V0+Vr if the potential barrier is lowered then the electric field islowered and so is the transition region. the same can be said backwards since the

transition region is narrower this might yield to lower barrier potential. if its wider itmight yield a higher potential barrier

------ Good Luck -----

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Due Date: 25/07/2008

Q.1. A germanium junction diode with n=2 has v=0.3v, conducts at 2mA.Find the voltage drop at i=10mA and i=20mA. Solution:

2 1 T 2 1

1

1

2 1 T 2 1

2

T

2

2

v - v = nV ln( i i ) v = 0.3V i = 2mA and n =2 thus v = v + nV ln( i i )for i = 10mA V = 25 mV=0.025 volts v = 0.3 + 2x0.025ln(10/2) v = 0.3 + 2x0.02

2

2

2

2

2

2

2

5ln(5) v = 0.3 + 2x0.025x1.60 v = 0.3 + 2x0.025ln(5) v = 0.38v

For i = 20mA v = 0.3 + 2x0.025ln(20/2) v = 0.3+ 0.05x2.30 v = 0.41v



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Q.2. For the circuit shown in the figure below, both diodes are identical, conducting 10mA at 0.7V and 50mA at 0.8V. Find the value of R for which V=80mV Sol.

D1, conduct 10mA at 0.7V and D2 50mA at 0.8V We know that lni2/i1 = V2-V1/nVT nVT = V2-V1 / ln (i2/i1) = 0.8-0.7 / ln5 = 0.0625 From the above figure i1+ i2 = 10mA -------- (I) V = Ri1 Where V=80mV=0.08v 0.08 = Ri1 -------- (II) i1 = Is e V1/0.0625 -------- (A)



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i2 = Is e V2/0.0625 ---------- (B) where, V2= V1 + 0.08 put this value in (B) i2 = Is e (V1 + 0.08)/0.0625 = Is eV1 /0.0625 (e0.08/0.0625) From (A) we have i2 = i1 3.59 From eq(I) we have, i1 + 3.59i1 = 10 i1 = 10/4.59 = 2.17mA From eq (II) we have, R = 80mV/ 2.17mA = 36.72Ω Q.3. A circuit shown in the figure below has values V1=150Vac, and RL= 10KΩ. What is the peak load current for the circuit?

Solution: The input voltage is given an rms value of 150 v. We convert it into peak voltage

1(pk) 1 (rms)

ac

1(pk) pk

V = V / 0.707

= 150V / 0.707V = 212.16 V

Now, the load voltage and current are found, after finding peak voltage as:

2(pk) 2 1 1(pk)

pk

2(pk) pk

V = N /N V

= (2/6)(212.16 V )

V = 70.72V

Finally, the load voltage is found as:

L(pk) 2(pk) F

L(pk) pk

V = V - V

= 70.7 - 0.7V = 70V

And peak load current will be:

L(pk) L(PK) L

pk

L(pk) pk

I = V /R

= 70 V /10k

I = 7mA



Assignment 7(Spring 2008)(Solution)

CIRCUIT THEORY (PHY301)MARKS: 30

Due Date: 06/08/2008

Q.1.(a) Determine the dc voltage and current values for the circuit shown in the figure below

also mention the units of each derived value.

Solution:We are given the 20Vac rated transformer, the peak secondary voltage is found as

V2(pk) = 20/0.707

=28.28VpkThe peak load voltage is now found as

VL(pk) = V2 – 1.4

VL(pk) = 28.28 – 1.4

VL(pk) =26.88VpkNow dc load voltage is found as

Vave = 2VL(pk)/Π=2(26.88)/3.14

Vave =17.12 VdcFinally the dc load current is found as

Iave = Vave/RL= 17.12/15k

Iave = 1.14mA

(b) What is the difference between filter and clipper?

FILTERS:The filters are used in power supplies to reduce the variations in the rectifieroutput signal. A clipping circuit consists of linear elements like resistors andnon-linear elements like junction diodes or transistor and capacitor.

CLIPPERS:The main function of the clippers are clip off a portion of the input signal withoutdistorting the remaining part of the alternating waveform. here are varieties of diodenetworks called clippers that have the ability to clip off a portion of the input signal withoutdistorting the remaining part of the alternating waveform. The half wave rectifier studied earlier isa simplest form of diode clipper. Depending on the orientation of the diode, the positive ornegative region of the input signal is “clipped” off. A clipping circuit consists of linearelements like resistors and non-linear elements like junction diodes or transistor, but itdoes not contain energy-storage elements like capacitors

Q.2 With the sine wave signal input of Fig (a), find the shape of the output signal VO inthe biased series clipper of Fig (b). What would happen if battery connections werereversed? Also draw the shape of out signal.

Solution:

First we will take the positive half signal of the input single. The diode become short as itis forward biased and no current will flow till Vg exceeds the opposing battery of 10V.By this, we can see that only the upper part of the positive single voltage will passesthrough the clipper circuit and appears as Vo across the R.The diagram of this flow is depicted below.

The negative signal will be clipped off in this circuit.The inverse configuration of the battery is shown below.

Reverse biased case:By reversing, during the positive half cycle of signal, the voltage across R will be sum ofsingle voltage and the battery voltage. This is shown below

The peak lower portion of the signal will be clipped off due to the battery during thenegative input of the half cycle.

(b) Describe the function of LED, voltage doubler and transistor.Function of LED:The main functions of LED’s are diodes that will emit light when biased properly.ight-emitting diode (LED) are widely used as indicator lights on electronicdevices and increasingly in higher power applications such as flashlights andarea lighting. An LED is usually a small area (less than 1 mm2) light source,

Voltage doublersA doubler is an electronic device that doubles the frequency of an input signal. Doublers are

occasionally used in wireless communications to obtain transmission frequencies higher thanthose normally possible for a given circuit design.

TransistorA transistor is a three terminal semiconductor device commonly used to amplify or switchelectronic signals. The transistor is the fundamental building block of computers, and all othermodern electronic devices. Some transistors are packaged individually but most are found inintegrated circuits.

.

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Circuit Theory - Solved Assignments - Semester Spring 2008