Assignment No 1 Solution

Q.1: In given circuit (a)Calculate the total current flow and the voltage drop across each resistor (b) Relative to point d, what will be the voltage at points, a, b and c

Sol: (a) We can see that 1Ω , 4Ω and 3Ω are in series so total resistance of circuit will be Rt=1+4+3 Rt=8Ω Given voltage is V=24v Total current can be found using Ohm’s law (V=IR) OR I=V/R=24/8 I=3A Now this same current 3A is flowing through all series resistance so Voltage across 1Ω will be V=IR1Ω

V=3x1=3v Voltage across 4Ω will be V=IR4Ω

V=3X4=12V Voltage across 3Ω will be V=IR3Ω

V= 3X3=9v (b) Voltage at point a relative to point d is Vda=24v Voltage at point b relative to point d is sum of voltage of 1Ω and 4Ω Vdb=15v Voltage at point c relative to point d is Vdc=3v

Q. 2: Find the equivalent resistance of the network given below. Write each step of the calculation to get maximum marks. Draw the circuit diagram of each step otherwise you will lose your marks.

Solution:

Starting from right side we see, at point a current will not pass through 5Ω , 14kΩ and 9kΩ but follow easy path (short circuit path) and reach to point a passing from point b, c and d so we neglect effect of 14, 5 and 9kΩ Now 8kΩ and 2kΩ are in parallel so 8x2/8+2=16/10=1.6kΩ 1.6kΩ and 2k are in series so 1.6+2=3.6kΩ At left side 6kΩ and 3k are in parallel, 6x3/6+3= 18/9=2kΩ

which is again in parallel of 2k 2x2/2+2=1kΩ 1kΩ is in series of 4kΩ so 1+4=5kΩ 5KΩ and 3.6kΩ are in parallel 5x3.6/5+3.6=18/8.6=2kΩ Q.3: Give brief answers

a) Conductors allow flow of current while insulator do not allow, WHY. Ans:Conductors, (metals) allow electricity to pass through them while insulators do not conduct electricity through them. This is because the conductors have free electrons which move in a specified direction on the application of some potential, causing the flow of current. Insulators don't have free electrons, because the electrons are tightly bound to the nucleus. In insulators the gap between the valance band and conduction band is large and requires large energy to produce free electrons. A conductor will allow current to pass through it easily because it has a low resistance. An insulator has a very high resistance and will allow very little current to flow through it.

b) Give two examples of DC voltage source and AC voltage source. Ans: Battery and dry cell are DC voltage source while AC generator and electricity from wapda grid station are AC voltage sources. c) In any circuit, state two requirements for producing current. Ans: For producing current in a circuit

1) Circuit must be closed circuit. 2) There should be a potential difference (voltage source) across the circuit.

Assignment No 2 Solution

Q.1. Find Unknown values of I and V for given diagrams.

Sol:For the network in fig( a) The circuit is open circuit , therefore no current will pass through the network. The voltage across the resistors must therefore be zero volts, as determined by ohm’s law (V=IR=OV) with the resistor simply acting as a connection from supply to the open circuit. The result is that the open circuit voltage will be V=10v as shown in fig. below. In fig. (b) The current IT will take the path of least resistance and since the short circuit condition at the end of network is the least resistance path, all the current will pass through short circuit. The voltage across the network will be same as that across the short circuit and will be zero volts, as shown in fig.

Q.2. First Identify and label each node in the network .Use nodal analysis to find Vx and the power of 6Ω resistance..

Sol. Fist of all we will identify and label each node,

From the above diagram we know that node V1 is directly connected with 15Volt voltage source so the voltage at Node V1 will be,

V1= 20Volts -------- (A) And node V3 = Vx -------- (B) Constraint equation v3 V2-V3=10Vx -------(C) Put the value of Vx from (B) into (C) we have,

V2-V3= 10V3 So, V2=11V3 ------- (D) Now we will write KCL equation at Super Node,

(V2 - V1)/2 + V2/4 + V3/5 + V3/10= 0 ----------- (E)

We know V1=20volts put in (E) we have,

(V2 - 20)/2 + V2/4 + V3/5 + V3/10= 0 10v2-200+5v2+4v3+2v3=0

15v2+6v3=200-------- (F)

Put the value of V2 from (D) into (F) we have,

15(11v3)+6v3=200 165V3+6v3=200

171v3=200 V3=1.16v

To find the power dissipated across 6Ω resistances. As 6Ω is directly connected to v1 and voltage at v1=20v so voltage drop at 6Ω is 20v Now Power dissipated at 6 ohm is P=V2/R

P=(20)2/6

P=66.66watt

Q.3. Find VAE and VCG in the circuit from all possible paths.

Solution:

Following path AEFGHA VAE +4v -12v+1v-10v=0 VAF=17v Now following AEDCBA VAF-2v-2v+6v-5v=0 VAF=3V Now we find VCG following path CGFEDC VCG+12V-4V-2V-2V VCG =-4V Now we find VCG following path CGHABC VCG+1V-10V+5V-6V=0 VCG=10V

Assignment No 3 Solution

Q # 1: Use Mesh analysis to find VO 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.

Solution

Q # 2: Use Nodal analysis to find IO in the given network. Identify and label each node 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.

Solution

Assignment No 4 Solution Q # 1: Find I o 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.

Sol. Applying principle of superposition, we will take effect of the sources one by one Only current source is acting 2kΩ and 2kΩ are parallel so our new circuit will be Using current division rule current through 2kwill be Io1=6x3/1+2+3=18/6=3mA Only Voltage source is acting

Now we apply voltage division rule V1.4k=4x1.4/2+1.4=5.6/3.4=1.64k As 1.4K is resultant of 2k and 5k so same voltage will drop across each there

I o2 =V/1.4=1.64/5=0.3mA Same current flows through 2k and 3k in series Io=Io1+Io2 Io= 3+0.3 Io=3.3mA Q # 2: Apply Source Transformation on the circuit given below to find VO. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.

Solution: From left side we see that 4mA and 4kΩ are in parallel and can be converted into voltage source V=IxR=4x4=16v In converted circuit 4kΩ will become in series of 16v voltage source as shown in fig. below. From right side we see 2mA current source in parallel of 6kΩ that can be converted into voltage source V=IXR=2X6=12V Circuit will adopt shape as Now we write KVL for loop 1 4kI1 -3+6kI1-12 -16=0 10kI1=31 I1=31/10=3.1mA

This 3.1mA current is flowing through 4kΩ so Vo can can be calculated by Ohm’s law Vo=3.1x4 Vo=12.4v Q # 3 Find VO in the network given below using Thévenin’s theorem. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.

Solution: In first step we remove load resistance RL that is 16Ω

In second step we find Vth at open terminal of circuit by nodal analysis, circuit will be as

KCL at Node V1 V1/10+(v1-v2)/5-4=0 V1+2V1-2V2=40 3V1-2V2=40 ……………(A) KCL at Node V1 (V2-V1)/5+(V2-10)/4 =0

4V2-4V1+5V2-50=0 -4V1+9V2=50…………………(B) Multiplying (A ) by 4 and (B) by 3 and adding we get V2=Vth=16.3v Third step: Calculating Rth

10 5 10 5 15is in series withΩ Ω= Ω + Ω = Ω

15 4 6015 4 3.15

15 4 19

Ω × ΩΩ Ω = = = ΩΩ + Ω

3.15 1 3.15 1 4.15is in series withΩ Ω= Ω + Ω = Ω

4.15thSo R = Ω

THEVENIN’S EQUIVALENT:

Vo = 16X16.3/(4.15+16) = 3V

Assignment No 5 Solution Q # 1: Find IO in the network given below using Norton’s theorem. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.

Solution:

First step:

LRemoving load resistor R 1 from the circuit.

Now the circuit will adopt the following shape.

k=

Second step: Calculating IN NNow short circuiting the circuit to find the value of I

After reshaping the circuits for using mesh analysis. We find In by loop analysis method

3 2 - 4 ----------(A)I I mA=∵

1

1 2 1

1 2 1

1 2

kvl for For mesh I

2 ( ) 3 12 0

2 2 3 12

5 2 12 ---------------(1)

k I I kI

kI kI kI

kI kI

− + − =− + =− =

2 3 2 1

2 3 2 1

3 2 1

3 2

2 2 1

2 2 1

2 1

For Super Mesh

3 2 2 ( ) 0

3 2 2 2 0

2 5 2 0

Putting 4 from (A) int the above equation

2k(4mA+ ) 5 2 0

8 2 5 2 0

7 2 8 ----------------------

kI kI k I I

kI kI kI kI

kI kI kI

I mA I

I kI kI

kI kI kI

kI kI

+ + − =+ + − =+ − =

= ++ − =

+ + − =− = −

1 2

1 2

2

2

2

2 3

-(2)

Multiply (1) by 2 and (2) by 5 we will get

10k 4 24 ----------------(1A)

10 35 40 ----------------(2A)

Adding (1A) and (2A) we will get

31k 16

16

310.51

(A)

- 4

I kI

kI kI

I

I

I mA

From

I I

− =− + = −

= −−=

= −

+ =

3

3

3

( 0.51 ) 4

4 0.51

3.48

mA

mA I mA

I

I mA

− − + == −=

Therefore 3.48NI mA= Third step: Calculating RN:Now we will find NR

We short circuit voltage source and open circuit current source

3 is in series with 2k so their combined effect is 5k.

Now the circuit will adopt the following shape

k

R32k

R23k

R15k

3k is parallel to 2k so their combined effect is

3k 2

3k 2

1 1 1

R || 3 2

2+3 =

65

=66

R ||5

=1.2k

k

k

R

R

= +

=

Now it will adopt the following shape

R21.2k

R15k

1.2k and 5k are in series so their combined effect is 6.2k. Therefore NR 6.2k=

Fourth step: Now reinserting the N L NR and R parallel to I

2

11 2

0

( ) *

6.2k =( )*3.48

1 6.26.2

=( )*3.487.2

=0.861*3.48mA

=2.99mA

Therefore I 3 app

k t

RI I

R R

mAk k

mA

mA

=+

+

=

Q # 2 Find VO in the network given below using Norton’s theorem. Show each step of calculation otherwise you will lose your marks. Draw and label the circuit diagram of each step and also mention the units of each derived value.

Solution: We want to calculate V

0 by using Norton’s theorem.we will follow these four steps .

First step: Replacing RL

with a short circuit to find IN.

Here Rl is 3kΩΩΩΩ

Second step:calculating IN

To calculate IN

We short circuit open terminal and find IN.

We apply super position method we will remove all circuits one by one i.e. after removing voltage source we will replace it with short circuit and current source with open circuit. Hint: Don’t remove all circuits simultaneously. Only current source is acting.

Due to short circuit all current will follow through the short circuit so I

N1 = 4mA --------------(A)

Only voltage source is acting.

4k is in parallel with 2k resistor which in return in series with 6k resistor .So total resistance. R = (4k||2k) +6k = 8/6 k +6k R = 7.33 k So I

N2 = 2/7.33k

IN2

= 0.27mA -------------- (B)

total I

N from both sources so from equation A and B we have

IN

= IN1

+ IN2

= 4mA + 0.27mA I

N = 4.27mA

Third step: Calculating R

N To calculate R

N we will short circuit all voltage sources and open circuit all current

sources.

4k is in parallel with 2k.The combined effect of these is in series with 2k. 4k||2k + 6k = 1.33 + 6k =7.33k =R

N Fourth Step:

After calculating IN

and RN

, re-inserting the load resistance RL

in the circuit in parallel RN

and

considering the IN

current source parallel with these two resistances. So our Norton

equivalent circuit will be.

by current divider rule we have I

0 = ( 4.27m)(7.33) x 1/10.33k

= 3mA By ohm’s Law we have V

0 = 3k x 3= 9 volts

Q # 3: Find the current ID and the diode voltage VD for the circuit in the fig. Assume

that the diode has a current of 1.5mA mA at a voltage of 0.7V. Solution: Here VDD=10V, R=2KΩ Diode voltage VD=0.7V ID can be found as ID = VDD – VD/R

= 10 – 0.7/2k = 4.65mA We know V2 – V1 = nVTln(I 2/I1)

log 10 x = (ln x)/ (ln 10) where ln10 =2.3

so 2.3 log 10 x = ln x

Now the equation of log to the base 10 form will be as V2 – V1 =2.3nVTlogI2/I1 For our case, 2.3 nVT = 0.1volts

V2 = V1 + 0.1logI2/I1 As V1 =0.7V, I1 =1.5mA and I2 =4.65mA putting these values above

V2 = 0.753V

So our required values of ID and VD are

ID =4.65mA and VD =0.75 V

Q # 4: (a) What do you understand by P type and N type semiconductor? What happens to pn junction when it is forward biased? Ans: P type is formed by doping trivalent element with silicon or germanium. Holes are majority in P type means current flow due to holes. N type is formed by doping pentavalent element with silicon or germanium. Electrons are majority carriers in N type means current flow due to electrons. The pn junction is excited by a constant current source supplying a current I in forward direction. The depletion layer narrows and barrier voltage decreases by V volts, which appears as an external voltage in the forward direction.

This current causes majority carriers to be supplied to both sides of the junction by the external circuit. Holes to the p material and electrons to the n material. These majority carriers will neutralize some of the uncovered, causing less charge to be stored in the depletion region. Thus the depletion layer narrows and the depletion barrier voltage reduces. This reduction in voltage cause more electrons to move from n side to p side and more holes to move from p side to n side. So that diffusion currents increases until equilibrium is achieved.

(b) Write the difference between characteristic of ideal diode model and practical diode model. Ans; The ideal diode behaves essentially as a switch:It is OFF means no current conducts when reverse biased and ON when forward biased, it has no resistance when forward biased,. The ideal diode characteristic is shown in Figure. v-i characteristic for ideal diode model is shown below

And ideal diode model is

Practical diode model

Practically we see there is a voltage drop of about 0.7 V across the diode (silicon; germanium is 0.3 V) when it is forward biased, and so ofenly we include this voltage drop in circuit analysis. Accordingly, the practical model is the ideal model with the addition of a voltage source in the forward bias model.

v-i characteristic for ideal diode model is shown below

And practical model is

Assignment 6 Solution Q # 1: Find the change in currents due to change of voltage by (a) 5mv, (b) -10mv, (c) 15mv using small signal model and exponential model of diode with n= 1 and biased at 2mA. Solution: For small signal model we have ∆v = rd ∆i

∆i = ∆v / rd

But rd = n VT / I

=1 x 25m/2m = 12.5ohms ∆i = ∆v / 12.5 Now from exponential model

i = Is e v / nVT

Also we know that

ID + ∆i = ID e ∆v/nVT

ID + ∆i = ID e

∆v/nVT

∆i = ID (e ∆v/nVT

- 1)

but ID = 2mA thus

∆i = 2 (e∆v/nVT

-1) For (a) ∆v = 5mV (1) ∆i = 5/12.5 = 0.4mA

(2) ∆i =2 (e5/25

– 1) =0.44mA (b) ∆v = -10mV (1) ∆i =-10/12.5

= -0.8mA (2) ∆i = 2(e-10/25 – 1) = -0.65mA (c) ∆v = 15mV

(1) ∆i = 15/12.5 = 1.2mA

(2) ∆i =2 (e15/25

– 1) =1.64mA Q # 2 A step up transformer has the turn ratio of 2:5, find the secondary voltage of transformer if primary voltage given is 240 Vac. Solution: Turn ratio of transformer is 2:5 which means N1=2 N2=5 V1=240v The turn ratio of the transformer is equal to the voltage ratio of the two components, so that

V2 = (N2/N1) V1 = 5/2x(240Vac) V2= 600Vac

Q # 3: Determine the dc load voltage and current values for the circuit shown in the figure below also mention the units of each derived value. SOLUTION:

The transformer has 20Vac input rated voltage, the peak secondary voltage is found as

V2(pk) = 20/0.707

=28.28Vpk

The peak load voltage is now found as

VL(pk) = V2 – 1.4

VL(pk) =26.88Vpk

The dc load voltage is found as Vave = 2VL(pk)/Π

=53.76/Π Vave =17.12 Vdc

Finally the dc load current is found as Iave = Vave/RL

= 17.12/10k Iave = 1.71mA

Q # 4: Find the waveform of output voltage VO for the following clipper circuit with given signal waveform at input. p

Solution:

During the positive half half cycle, D1 will conduct but D2 will act as an open circuit

means does not conduct current. But value of Vo cannot exceed 7v because point C and

D are electrically connected across 7v battery since D1 is shorted. Hence signal voltage

above +7v level would be clipped off as shown in fig below.

During the negative half half cycle ,D1 is open but D2 conduct current. But value of Vo

cannot exceed 5v because point C and D are electrically connected across 5v battery

since D2 is shorted. Hence signal voltage beyond 5v level would be clipped off as shown

in fig