30
Assignment 1(Fall 2007) (Solution) CIRCUIT THEORY (PHY301) MARKS: 35 Due Date: 12/10/2007 Q.1. Determine the power that is absorbed or supplied by the circuit elements in figure below. Sol: Power of 24v voltage source: As we see current is flowing out of the +ve terminal of 24v voltage source so according to passive convention power is supplied by this element, its value is P=VXI=24X2=48w Power of element 1: we see current is flowing into the +ve of element 1 so according to passive convention power is absorbed by this element, its value is P=VXI=10X2=20w Power of element 2: Current is flowing into the +ve of element 2 so according to passive convention power is absorbed by this element, its value is P=VXI=14X2=28w Power supplied=power absorbed 48v=20+28 =48w

Circuit Theory - Solved Assignments - Semester Fall 2007

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Page 1: Circuit Theory - Solved Assignments - Semester Fall 2007

Assignment 1(Fall 2007) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 35

Due Date: 12/10/2007

Q.1. Determine the power that is absorbed or supplied by the circuit elements in figure below.

Sol:

Power of 24v voltage source: As we see current is flowing out of the +ve terminal of 24v voltage source so according to passive convention power is supplied by this element, its value is

P=VXI=24X2=48w

Power of element 1: we see current is flowing into the +ve of element 1 so according to passive convention power is absorbed by this element, its value is

P=VXI=10X2=20w

Power of element 2: Current is flowing into the +ve of element 2 so according to passive convention power is absorbed by this element, its value is

P=VXI=14X2=28w

Power supplied=power absorbed

48v=20+28

=48w

Page 2: Circuit Theory - Solved Assignments - Semester Fall 2007

Q.2.

In the network given below find the voltage Vab .and Vs

Sol:

40v is dropped across 20Ω , therefore from Ohm’s law current I2 will be

I2=V/R=40/20=2A

As we know same current flows in series so same current I2 flowes through 10Ω.

Vab=IXR

=2X(10+20) ( As 60Ω and 30Ω are in parallel, same voltage drops)

Vab =60v

I3=Vab/60=60/60=1A

We know entering current =leaving current so

I1=I2+I3

=2+1

I1 =3A (that is current flowing through 5kΩ) therefore

V5k=I1XR=3AX5KΩ=15Kv OR =15000V

Vs=V5k+Vab=15000v+60=15060v

Page 3: Circuit Theory - Solved Assignments - Semester Fall 2007

Q.3.

For the circuit shown in the figure below, all the resistors are given in K Ohms; Find the total resistance RT in the following circuits. Draw the circuit diagram of each step otherwise you will lose your marks. Draw the circuit diagram of each step 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:

Starting from right side we see, at point a current will not pass through R7(6Ω) but follow easy path (short circuit path) and reach to point b.At point b again due to short circuit current will not pass through R6 and R2 but follow short circuit path and thus passing through 7Ω completes its path to reach voltage source V, so effect of 23Ω , 6Ω(R6) and 6Ω(R7) is neglected. Therefore circuit adopts the form as From left we see that R1 and R3 are in parallel so 8x8/8+8=64/16=4Ω 4Ω and 2Ω are in series so 4+2=6Ω Here 7Ω and 6Ω are in parallel so 7x6/7+6=42/13=3.23Ω Hence Rt=3.23Ω

Page 4: Circuit Theory - Solved Assignments - Semester Fall 2007

Assignment 2(Fall 2007)

(Solution) CIRCUIT THEORY (PHY301)

MARKS: 40

Due Date: 29/10/2007

Q.1. For the network given below

(a) Find the total resistance and conductance.

(b) Determine Is and the current through each parallel branch.

Sol: (a) We see here that all resistances are in parallel So 1/Rt = 1/R11/R21/R3 Put the values 1/Rt =1/101/51/4

1/Rt = (245)/20 1/Rt = 11/20 Rt= 20/11 Rt = 1.81kΩ Now To calculate Conductance We know conductance is the reciprocal of resistance, means Conductance = 1/Rt = 1/1.81KΩ = 0.55 mili Siemens

Page 5: Circuit Theory - Solved Assignments - Semester Fall 2007

(b) As all the resistors are in parallel so same 20v battery voltage are applied across the all resistors.

11

10

VI

k

1

20

10I

k

1 2I mA

22

5

vI

k

2

20

5I

k

2 4I mA

33

4

vI

k

3

20

4I

k

3 5I mA

Apply the KCL Rule, we get

1 2 3SI I I I

2 4 5

11

S

S

I mA mA mA

I mA

Q.2. Use Nodal analysis to find Voltage drop across 10Ω resistance in the network

given below. Identify and label each node 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.

Page 6: Circuit Theory - Solved Assignments - Semester Fall 2007

Sol: Labeling the nodes

1 1 2

1 1 2

1 2

1 2

2 2 2 1

2 2

KCL for node 1:

V V V10 0

10 20

2V V V10

20

3V V10

20

3V V 200........................................................ 1

V V 12 V V0

2 4 20

2V V 12

4

Writing KCL for the two nodes;

KCL for node 2:

2 1

2 2 1

2 2 1

2 2 1

2 1

V V0

20

5 3V 12 +V V 0

15V 60 V V 0

15V +V -V 60

16V -V 60..............................................(2)

1 2

2

1

1

Multiplying equ. (1) by 16 and adding in (2)

48V 16V 3200

-V1+16V =60

-----------------------------------

47V 3260

3260 V

47

V1 6

9.36v

That is the voltage drop across 10 Ω

Page 7: Circuit Theory - Solved Assignments - Semester Fall 2007

Q.3. Use nodal analysis to find VO in the network given below. Label each node

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: Labeling fig.

6k ohm resistance is short circuited so role of 6kΩ is neglected.

3 4 0v v

Because they are connected to ground directly.

As 10v voltage source involves between two nodes so Making v1 and v2 a super node

Equation for super node

31 2 2

3 3 34 10 0

4 10 3 10 6 10

v v v

1 2 2 44 3 6

v v v

Page 8: Circuit Theory - Solved Assignments - Semester Fall 2007

1 2212 1212

4(12)4 3 6

v vv

1 2 23 4 2 48v v v

1 23 6 48v v -----------------------------------(1)

Also the constraint or coupling equation is

1 2 10v v ---------------------------------(2)

Multiplying (2) by 6 and adding in (1)

1 23 6 48v v

1 26 6 60v v

19 108v v

1 12v v

From the given figure

1 4 0v v v

Also

4 0v

So

1 0v v

0 12v v

Page 9: Circuit Theory - Solved Assignments - Semester Fall 2007

Q.4.

(a) How much resistance is required to limit the current to 1.5 mA, if a 6v battery is

connected across the resistance?

Ans: We are given here

V=6v

I=1.5mA OR 1.5x10-3A

By ohm’s law V=IR so

R=V/I

R= 6/1.5x10-3

R=4KΩ

(b) Give your arguments about the statement that every short circuit is a close circuit but

every close circuit is not a short circuit.

Ans: It is necessary condition for a short circuit that it should be closed as current

cannot pass through open circuit. An electricl device will function only if the circuit is

closed. ie. current is allowed to pass through it. In the circuit there must be a energy

consuming device.

But before reaching any device if the circuit is completed (closed) heavy current will be

flowing through the circuit This is called "short circuit"

If the circuit is completed It is closed circuit whether it is passing through a device or not.

If the circuit is completed without a device it can be called a "short circuit"

……..Good Luck………

Page 10: Circuit Theory - Solved Assignments - Semester Fall 2007

Assignment 3 (Fall 2007)

(SOLUTION) CIRCUIT THEORY (PHY301)

MARKS: 30

Due Date: 09/11/2007

Q.1. For the network given below Find V1 and V2

Sol: We use Loop method and assign loop 1 and 2 for the given circuit.

Now we write KVL, and as from KVL, the sum of voltage around a closed loop is = 0

Therefore, writing KVL for loop 1

3 + V1-6 = 0

V1 = 3V

Writing KVL for loop 2 on the right:

-10 + V2 - V1 = 0

By putting the value of V1 we get:

-10 + V2 - 3 = 0

V2 = 13 V

Page 11: Circuit Theory - Solved Assignments - Semester Fall 2007

Q.2.

You are given the network below. Use any method you desire to find Vda . Show your

complete work. Label circuit diagram properly otherwise you will loose marks..

SOL:

We assume that current I is passing through close path

Writing KVL for closed path abcde:

-30 + 8I + 12I – 70 + 5I + 10I + 50 + 15I= 0

-50 + 50I=0

50I=50 or I = 1A ------------- (i)

So the current flowing in the circuit is 1A.

Now we’ll calculate the Voltage between a and d. (Vda) following the path abcd

-30 + 8I + 12I – 70 + 5I + Vda=0 => -100 + 25I + Vda=0

Vda = 100 – 25I --------------- (ii) Put the value of I from (i) into (ii)

Vda= 100 – 25 (1)

Vda= 100 – 25

Vda= 75v

We get same result if follow dea path

Writing kvl for dea path

10I+ 50+15I+Vad=0

10+50+15+Vad=0

75+Vad=0

Vad=-75 or Vda=75v

Page 12: Circuit Theory - Solved Assignments - Semester Fall 2007

Q.3. Use loop analysis to find current Io through 10Ω resistance in the network given

below. Identify and label each loop 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: Labeling circuit for loop currents

Here I3=4A

Writing KVL for loop I1:

6I1 + 2 (I1 - I2)-10 = 0 => 6I1 + 2I1 - 2I2-10 =0

8I1 - 2I2 = 10 ----------------- (I)

KVL for loop I2:

10I2 + 4(I2 – I3)+ 2 (I2 - I1) = 0 => 16I2-2I1-4I3=0 putting I3=4

-2I1 + 16I2 = 16 ----------------- (II)

Multiplying (II) by 4 and adding in (I)

8I1 - 2I2 = 10

-8I1+64I2=64

62I2=74

I2=74/62

I2 =1.19

As Io=I2

so Io=1.19A

Page 13: Circuit Theory - Solved Assignments - Semester Fall 2007

Assignment 4(Fall 2007) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 30

Due Date: 28/12/2007

Q.1. Use nodal analysis to find IO in the given network. Identify and label each node otherwise you will lose your marks. Label circuit diagram properly. Write each step of the calculation to get maximum marks.

Sol:

Labeling circuit diagram for nodes

Writing KCL for node V1 V1/3+(V1-2)/2+(V1-V2)/1=0 2V1+3V1-6+6V1-6V2=0 11V1-6V2=6………………(I) Writing KCL for node V2 V2/5+(V2-V1)/1-2=0 V2+5V2-5V1-10=0 -5V1+6V2=10…………….(II) Adding (I) and (II) 11V1-6V2=6 -5V1+6V2=10 6V1=16

Page 14: Circuit Theory - Solved Assignments - Semester Fall 2007

V1=2.66v Putting V1 in (I) 11(2.66)+6V2=6 29.33-6V2=6 V2=23.33/6 V2=3.88v Here Io=V1-V2/1 =2.66-3.88/1 Io =-1.22A

Q.2. Use Mesh analysis to find voltage Vo in the given network. Draw and labeled complete circuit diagram otherwise you will lose your marks. Sol: Labeling fig. for mesh KVL for loop 1

I1 = -5mA KVL for loop 2 2(I2 – I1) - 15 + 1(I2-I3) = 0 2I2 – 2I1 - 15 + I2 – I3 = 0 -2I1 + 3I2 – I3 = 15 Putting I1 in Eq. -2(-5) + 3I2 – I3 = 15 10 + 3I2 – I3 = 15

Page 15: Circuit Theory - Solved Assignments - Semester Fall 2007

3I2 – I3 = 15 – 10 3I2 – I3 = 5…………………(A) OR I2=5+I3/3 KVL for loop 3 15 + 1(I3 – I2) + 3(I3 – I4) = 0 15 + I3 – I2 + 3I3 – 3I4 = 0 -I2 + 4I3 – 3I4 = -15 Putting I2 value and simplifying we get 11I3-9I4=-40 ……………(B) KVL for loop 4 4I4 +10 + 3(I4 – I3) = 0 4I4 + 3I4 – 3I3 = -10

-3I3 +7I4 = -10…………(C) Multiplying (B) by 7 and (C) by 9 and adding we get I3=-7.4mA so I2=-0.8mA Therefore VO=(I2-I3) 1K VO=(-0.8+7.4)10-3 X103

VO=6.6v

Q.3 First Define all nodes and Identify and label each Mesh in the network. Use Mesh analysis to find Current IO and Voltage Vo in the given network. 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.

Page 16: Circuit Theory - Solved Assignments - Semester Fall 2007

Sol: Labeling the figure for mesh and super mesh KVL equation for loop 1 is 10I1+20(I1-I2) + 4V0 -50=0 As V0=30I3 so putting V0 value in above eq. 10I1+20I1-20I2+4(30I3)=50 30I1-20I2+120I3=50……….…..(A) There is a current controlled current source between mesh 2 and 3, so Mesh 2 and Mesh 3 form a Super Mesh:

Writing KVL for super mesh 40I2+30I3+20(I2-I1)-4V0 –120=0 40I2+30I3+20I2-20I1-4(30I3)=120 -20I1+60I2-90I3=120……………..(B) Constraint equation is I3-I2=2I0 As I0=I2 So I3-I2=2I2 3I2-I3=0……………………….(C) Multiplying eq (A) by 2 and (B) by 3 and adding we get 140I2-30I3=460 …………….(D) Multiplying (C) by 30 and subtracting from (D) 140I2-30I3=460 -90I2+30I3=0 50I2=460 I2=9.2A =I0 Putting I2 in (C) I3=27.6A V0=30I3 =30(27.6) V0=828v

Page 17: Circuit Theory - Solved Assignments - Semester Fall 2007

Assignment 5(Fall 2007) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 35

Due Date: 17/01/2008

Q.1

Using source transformation find VO in the following network. Draw and label each circuit diagram, otherwise you will lose your marks. Write each step of calculation and also mention the units of each derived values.

Solution: From left side we see 8k is in parallel with 3mA source, so it can be converted into a voltage source, using ohms law

Ω

(3 )(8 ) 24= = Ω =V IR mA k VOur modified circuit will be: In the above circuit resistor is in series with 8kΩ 4kΩ . we add them as 8k+4k=12k

Page 18: Circuit Theory - Solved Assignments - Semester Fall 2007

24V is in series with 12 resistor so it can be converted into a current source of value: Ωk24 2

12= = =

ΩV VI mAR k

Circuit will be

In the above circuit two current sources are parallel to each other, so current sources will add up to give value:

2 2 4= + =I mA mA mA So the circuit adopts the form as : Now 12 resistor is in parallel with kΩ 6kΩ resistor so;

12 612 || 6 412 6

k kk k kk k×

Ω Ω = =+

Ω

V

4k resistor is in parallel with 4mA source, so it can be converted into a voltage source.

Ω

(4 )(4 ) 16= = Ω =V IR mA k VOur modified circuit will be: Positive terminal of 16V battery is connected with the positive terminal of 12V battery and negative terminal of 16V battery is connected with the negative terminal of 12V battery. Net voltage will be

16 12 4= − =V V V The circuit would be: Applying voltage division rule:

( )

1o

1 2

o

V

4 44 41 6 8

V = 2 V

= ×+

= ×+

=

sR V

R Rk

k k

Page 19: Circuit Theory - Solved Assignments - Semester Fall 2007

Q.2. Find the Vo in the network below using Thevenin’s Theorem. 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.

…………Good Luck………..

Sol: Step 1: Removing load resistance Here 2k is load resistance across which we have to find Vo. Step 2: Calculating Vth

We use nodal analysis method to determine Vth. At node 1 V1-V3/2k = -2mA + 1mA therefore V1-V3 = -2 ------------- (1) At node 2 V2/1k + V2-V3/1k - 2mA=0 2V2 –V3=2 ---------------- (2) At node 3 V3-V2/1k + V3-V1/2k+ V3-Vx/1k =4mA 5V3 -2V2-V1-2Vx=8 ----------------- (3) At node x

Page 20: Circuit Theory - Solved Assignments - Semester Fall 2007

Vx-V3/1k + 1mA =0 Vx –V3 = -1------------ (4) Writing equitation (3) in terms of V3 and Vx we have from (1) and (2) 5V3 -2-V3-V3+2-2Vx = 8 3V3 -2Vx = 8 Putting value of V3 from (4) 3Vx +3-2Vx=8 Vx =5V so Voc= Vx =5V Third step : Calculating Rth

by open circuiting the current source Here 2kΩ is not in close path but has open terminal so We will ignore its effect in the circuit Therefore Rth will be sum of series 1k+1k+1k=3kΩ Rth=3kΩ

Fourth step. Now Vo can be found reinserting load resistance 2kΩ in series of Rth (3kΩ) and Vth Using voltage division rule, Vo will be Vo= 5 2

3 2x+

=10/5

V0=2v

…………Good Luck…………

Page 21: Circuit Theory - Solved Assignments - Semester Fall 2007

Assignment 6(Fall 2007) (Solution)

CIRCUIT THEORY (PHY301) MARKS: 40

Due Date: 31/01/2008

Q.1 Find VO in the network given below using Thevenin’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: Removing Load resistance We remove load resistance 2k Second resistance: Finding Vth To find Vth we use node analysis method so we assign nodes V1 and V2 Constraint equation is V2-V1=2kIx……………(A) As Ix= 5-V1 Putting this value of Ix in (A) 2k V2-V1=2k(5-V1)/2k V2-V1=5-V1 V2=5v Actually V2=Vth so Vth=5v

Page 22: Circuit Theory - Solved Assignments - Semester Fall 2007

Third step: Finding Rth Because dependent voltage source involves in the circuit so working with dependent sources is different from working with independent sources while applying Thevenin’s theorem . For dependent source while calculating R

th we will short circuit the open terminals of the Thevenin circuit and

will calculate the Isc

and then divide Vth

with Isc

to calculate Rth

.

Using nodal method

Isc=V2/1k V can be calculated as in second step above which is V2=5v so 2 Isc=5/1k Isc=5mA Now Rth will be Rth=Vth/Isc =5/5mA Rth =IKΩ

Fourth step: Reinserting Load resistance 2k in series of Rth and Vth

Now Vo can be calculated using voltage division rule. Vo=2x5/2+1=10/3 Vo=3.33v

Page 23: Circuit Theory - Solved Assignments - Semester Fall 2007

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 unit of each derived value.

Solution: First step: Here 3kΩ is load resistance, replacing RL with a short circuit to find IN.

Second step: Finding INWe find IN using loop method.

Here I1=4mA Writing KVL for loop 2 6kI2-5+2k(I2-I1)=0 6kI2-5+2kI2-2kI1=0 Putting I1 value 6kI2-5+2kI2-8=0 8kI2-13=0 I2=13/8k I2=1.62mA Since I2=IN So IN=1.62mA Third step: Calculating Rth To calculate Rth we short circuit voltage source and open circuit current source of 4mA. Since 6kΩ (To the left of circuit) becomes open due to open circuiting, we ignore its effect. Now 2k and 6k are in series their sum is 2k+6k=8kΩ so Rth=8kΩ

Page 24: Circuit Theory - Solved Assignments - Semester Fall 2007

Fourth step: Re-inserting Load resistance RL in the circuit in parallel of Rth and IN. First we find Io using current division rule. Io=8x1.62/8+3=12.96/11 Io=1.17mA Now Vo=Iox3k =1.17mAx3k Vo=3.53v

Q.3 Below are given three diodes (a),(b),(c) with different polarities. Describe reverse/forward biased condition of each with justification. Answers: Fig(a) In fig(a) diode is in reverse biased condition because +ve terminal of voltage source is connected to the cathode of the diode through a resistance . The anode terminal of the diode is grounded.. Now the diode is in reversed biased condition so, it will act as an open circuit. Fig (b) In fig (b) diode is also in reversed biased condition because -ve terminal of voltage source is connected to the anode of the diode through a resistance. The cathode terminal of the diode is grounded. Now the diode is in reversed biased condition so, it will act as an open circuit. Fig(c) In fig(c) diode is in forward biased condition because -ve terminal of voltage source is connected to the cathode of the diode through a resistance. The anode terminal of the diode is grounded.. Now the diode is in forward biased condition so, it will act as closed circuit.

Page 25: Circuit Theory - Solved Assignments - Semester Fall 2007

Assignment 7(Fall 2007)

(Solution)

CIRCUIT THEORY (PHY301)

MARKS: 25

Due Date: 19/02/2008

Q.1.

Assume that the diode is germanium diode. Determine power dissipated through

3kΩ resistance. Also find current and voltage of resistance, considering diode to be

ideal. Show each step of calculation otherwise you will lose your marks.

Sol:

Here 10v is the source voltage (VS). As the diode is forward biased , so some of

voltage drops across diode for its farward biasing(VF) and remaining across

3KΩ (VR).Therefor

VS=VF+VR

To find power dissipated through 3kΩ, first of all we determine VR so from above

VR=VS-VF For germanium VF=0.3v

VR=10-0.3

VR=9.7v

As P=VI=V2/R

P=V2/R

P=(9.7)2/3k

P=31.36mw

Now if we consider this diode to be ideal then no voltage drop will occur

Across it (VF=0), then we have

VS=VR

VR=10v

I=VR/R

I=10/3

I=3.33mMA

Formatted: Font: Bold

Page 26: Circuit Theory - Solved Assignments - Semester Fall 2007

Q.2

Determine the dc voltage and current values for the circuit shown in the figure

below.Also find the diode Peak inverse voltage. Mention the units of each derived

value.

Sol.

Here we are given 20Vac rated transformer, 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)

=53.77/Π

Vave

=17.12 Vdc

Finally the dc load current is found as

Iave

= Vave

/RL

= 17.12/10k

Iave

= 1.71mA

Diode peak inverse voltage is

PIV= VL(pk)

+0.7v

=27.58v

Page 27: Circuit Theory - Solved Assignments - Semester Fall 2007

Q.3: Consider a diode with n=1 which biased at 2mA.Find the change in currents as a result of

changing the voltage by (a) -10mv (b) +5mv (c) +10mv , using small signal model and

exponential model.

Sol:

For small signal model we have

Δv = rd Δi

Δi = Δv / rd

But

rd = n VT / I

=1 x 25m/2m

= 12.5ohms

Δi = Δv / 12.5

(a) Changing voltage at -10mv

Δi = -10 / 12.5

= -0.8mA

(b) For Δv=5mv

Δi = 5 / 12.5

=0.4mA

(c) For Δv=10mv

Δi = 10 / 12.5

=0.8mA

Now from exponential model i = Is e v / nVT

we know that

ID + Δi = ID e Δv/nVT

ID + Δi = ID e Δv/nVT

Δi = ID (e Δv/nVT - 1)

We are given

ID = 2mA thus

Δi = 2(eΔv

/nVT -1)

(a) For Δv=-10mv Δi = 2(eΔv

/nVT -1)

Δi = 2(e-10

/25 -1)

=2(-0.329)

=-0.659mA

(b) For Δv= 5mv Δi = 2(e5/25 -1)

Page 28: Circuit Theory - Solved Assignments - Semester Fall 2007

= 2(1.22-1)

=0.44mA

(c) For Δv= 10mv Δi = 2(e10

/25 -1)

=0.98mA

Page 29: Circuit Theory - Solved Assignments - Semester Fall 2007

Assignment 8(Fall 2007)

(Solution)

CIRCUIT THEORY (PHY301)

MARKS: 25

Due Date: 29/02/2008

Q.1. Draw the schematic diagrams for half wave and full wave rectifier and their typical

output waveform. Also write formulae to determine

i) Peak load voltage ii) DC load current iii)PIV , for both rectifiers.

Answer:

Half wave Rectifier:

(Input waveform) (Half wave rectifier) (Out put waveform)

i). Peak load voltage:

VL(pk) = V2(pk) – VF

ii). DC load current:

VL(pk) / RL

iii). PIV:

V2(pk)

Full wave rectifier:

11 1

Page 30: Circuit Theory - Solved Assignments - Semester Fall 2007

i). Peak load voltage:

VL(pk)=V2(pk) / 2 - VF

ii). DC load current: I

L(pk)= V

L(pk)/R

L

iii). PIV:

2VL(pk) + 0.7V

Q.2. For each of the circuits shown in the figure below, find the emitter, base, and collector

voltages and currents. Use 20, but assume |VBE| = 0.3V independent of current

level.

SOL:

From fig (A) have VB = 0V

VE = VB+0.3 =0.3V

IE= (5-VE) / 1.5 = (5- 0.3)/1.5 = 3.13mA

IC = ά IE = 20/21 x 3.13mA = 2.98mA

VC = -5 + 2xIC = -5 + 5.96 = 0.96V

IB = IC/ = 2.98 / 20 = 0.149mA

From fig (B) we have

VB = 4V

VE = VB+0.3 =4.3V

IE= (12-VE) / 2 = (12-4.3)/2 = 3.85mA

IC = ά IE = 20/21 x 3.85mA = 3.66mA

VC = IC x 5= 18.33V

IB = IC/ = 3.66/ 20 = 0.183mA