Circuit Theory - Solved Assignments - Spring 2009

Assignment 1(Spring 2009)(Solution)

CIRCUIT THEORY (PHY301)MARKS: 30

Due Date: 26/03/2009

Question #1:Find Equivalent resistance Rt of given circuit and also find total current flowing in

circuit .Draw the circuit diagram of each step otherwise you will lose your marks. Writeeach step of the calculation to get maximum marks and also mention the units of eachderived value.

Solution:From left side we seeR2 & R4 are in series so (1+2) = 3kΩ

So circuit will become as

R2 & R3 are in Parallel so (3*2) / (3+2) = 1.2KΩ

R1 & R2 are in Series so (4+1.2) = 5.2KΩAnd new form of circuitwill be as

Right sideNow starting from Right side we see R6 andR7 are in parallelSo equivalent of 6k and 6k is

6k 6k= (6 6)/ (6+6)=36/12=3k

The resistors 7K , 3K and 3K are in series so they will be combined as=7k+3k+3k=13k

5.2KΩ and 13kΩ are parallel resistances so the equivalent value is13x5.2/13.5.2=3.71KΩ

So the total resistance of the circuit is 3.7kΩ, applied voltage is 25 VSo using Ohm’s law the current of the circuit is given by the formula.I=V/RThat isI=25V/3.7 KΩI=6.7mA

Question #2:Calculate the value of source voltage Vs. Draw the circuit diagram of each step otherwiseyou will lose your marks. Write each step of the calculation to get maximum marks andalso mention the units of each derived value

Solution;It is given that voltage drop across 6Ω is 12v so by Ohm’s law current through it will be

I=V/R=12/6=2A

As 2Ω is in series of 6Ω so same 2A current flow through 2ΩSo voltage drop across 2Ω will be

V=2AX2ΩV=4v

As 2Ω is in series of 6Ω, their sum is 2Ω+6Ω=8ΩAnd total voltage drop across 8Ω due to sum of 12v and4v is

V=12v+4vV=16v

Circuit can be redraw as

Here 8Ω and 8Ω are in parallel, their equivalent is 8x8/8+8=64/16=4ΩAnd same voltage drop across parallel equivalent so we can redraw above circuit as

Now using voltage divider rule we have

V4Ω=

16v=

Vs=

Vs= 32vQuestion #3:Briefly answer the following questions

i. What will be the Atomic Number of element if 12 electrons revolve in shells?Ans: Atomic Number of element will be 12

ii. What differences and similarities are there between short circuit and open circuit?Ans: Open Circuit:-When any part of the path is broken, the circuit is open because thereis no continuity in the conducting path.

The resistance of an open circuit is infinitely high. The result is no current in the open circuit. An open occurs when there is a physical break anywhere in the circuit or indeed

within an individual component forming part of the circuit. Due to this physicalbreak, an open means that no current is able to flow, thereby either being unableto reach the load or return through the earth path to the battery.

Short Circuit:-In this case, the voltage source has a closed path across its terminals,but the resistance is practically zero.

The result is too much current in the short circuit. Usually, the short circuit is a bypass across the load resistance.A short, meanwhile, occurs when a component or the wiring is damaged, forexample, as a result of the insulation chaffing, and the electrical current in the circuitbeing able to find a new path back to the battery. Typically, with automotive wiringthe short will occur when exposed wire(s) are able to touch each other, bodywork or acomponent, allowing the current to go to earth rather than take the long way, i.e. itsintended path, back to the batteryAns:

iii. What will be the power dissipation of 5kΩ resistor if 2A current flow through it.Ans: As R=5kΩ and I=2A

Power P=VXI=I2RP=(2)25P=20KW

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Assignment 2(Spring 2009)

(Solution) CIRCUIT THEORY (PHY301)

MARKS: 40

Due Date: 13/04/2009

Q.1

You are given the circuit shown in the figure below. Determine the current I. Does not

use mesh analysis in your solution. Draw diagram where necessary.

Sol: As there is a short circuit wire (direct wire without any resistance) in the circuit and we know current always follow easy and short circuit path to flow. So reaching current at point A it will flow through short circuit path and will not pass through 15Ω(thus also through 40 , 40Ω) and 35Ω.so we ignore the effect of these resistances in labeled box.

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So our remaining circuit will be as drawn below

Now looking at right side as 2Ω & 4Ω are in series with each other the

circuit becomes;

Req=2+4 =6Ω

+

-10 V

7 Ω

12 Ω 30 Ω

20 Ω

6 Ω

Now as 6Ω & 4Ω are parallel to each other the circuit becomes;

6 x 20 120 4.615

6 20 26eq eq

R R= ⇒ = = Ω+

+

-10 V

7 Ω

12 Ω 30 Ω 4.615 Ω

Now as 4.615Ω & 30Ω are parallel to each other the circuit becomes;

30 x 4.615 138.45 4

30 4.615 34.615eq eq

R R= ⇒ = = Ω+

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+

-10 V

7 Ω

12 Ω4 Ω

Again here 4Ω & 12Ω are parallel to each other the circuit becomes;

12 x 4 48 3

12 4 16eq eq

R R= ⇒ = = Ω+

Now as 3Ω & 7Ω are in series with each other so we add them

Req=3+7 =10Ω

Now

The total current flowing in the circuit van be calculated using Ohm’s law

Here

10

10

10 1.0

10

0

t

t

t

t

t

V v

R

VI

RA

=

= Ω

= = =

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Q.2 First label each node in the network .Use nodal analysis to find Vx . Write each step of

calculation otherwise you will loose your marks. Also mention the units of each derived

value.

Solution:

First of all we label and identify nodes in network.

In the above circuit, Vx = V4

As there exists a voltage source between node V1 and V2 we assign it super node 1.

Constraint equation for super node 1 is

V2-V1 = 100 …………….(1)

Also there exits voltage source between node V3 and V4 we assign it super node 2.

Constraint equation for super node 2 is

V3-V4 = 150 …………….(2)

By redrawing, the given circuit is

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Applying KCL at super node 1

Now by applying the KCL at super node 2:

By solving equations 1, 2, 3, and 4, we get approximate values as

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V4 = Vx = 60.6312V

Q.3

You are given the circuit in the figure

Find VA and VB by using Nodal Analysis.

a) Determine the current Ix.

b) Determine the power supplied by the dependent voltage source.

c) Determine the power supplied by the independent current source.

d) Determine the power supplied by the independent voltage source.

e) Determine the power absorbed by the 6 ΩΩΩΩ resistor.

f) Determine the power absorbed by the 3 ΩΩΩΩ resistor.

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Sol:

At node 1: -----(1)

At node 2: ------(2)

At node 2, we have super node: -------(3)

At node 1: -------(4)

substitute from (4) into (2), we get ------(5)

Now sustitute (3) into (5), we get -------(6)

(1) simplifies to ----------(7)

(6) simplifies to ----------(8)

On solving (7) & (8), we get

(a)

.

(b)

Power supplied by dependent voltage soure:

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From (4),

.

(c)

Power supplied by independent current source:

.

(d)

Power supplied by independent voltage source:

.

(e)

Power absorbed by resistor:

.

(f)

Power absorbed by resistor:

.

-:----------------------------------------------:-

Q.4

First Identify and label each mesh in the network .Use Mesh analysis to find all mesh

currents of circuit. Write each step of calculation otherwise you will loose your marks

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Solution:

We assign meshes to circuit

Writing KVL equation for mesh1:

3I1 + 1(I1 – I2) = 10

3I1 + I1 – I2 = 10

4I1 – I2 = 10…………………….. (A)

Writing KVL equation for mesh2:

5I2 + 2(I2 – I3) + 12 + 1 (I2 – I1) = 0

5I2 + 2I2 – 2I3 + 12 + I2 – I1 = 0

- 2I3 + 8I2 – I1 + 12 = 0 ………(B)

For mesh 3:

In mesh 3 we are given independent current value so

I3 = -3A (-ve sign with 3A is because direction of I3 mesh and given 3A are in

opposite as arrow shows)

Put I3 in (B)

-2(-3) + 8I2 – I1 + 12 = 0

6 + 8I2 – I1 + 12 = 0

8I2 – I1 = -18 (C)

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Multiply (C) by 4

32I2 – 4I1 = -72 ……………..(D)

Solving (A) and (D)

-I2 + 4I1 = 10

32I2 – 4I1 = -72

31I2 = -62

I2 = -2A Put in (A)

4I1 – (-2) = 10

4I1 + 2 = 10

4I1 = 8

I1 = 2A

So our required mesh values are

I1 = 2A , I2 = -2A and I3 = -3A

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

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Assignment 3(Spring 2009)

(Solution) Circuit Theory (PHY301)

Mark: 40

Due Date: 29/05/2009

Q.1: Transform the following network to a single voltage source. 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:

2mA and 6kΩ are in parallel so convent into voltage source

V=IxR=6X2=12V

Here 2kΩ, 4kΩ and 6kΩ are in series so 2+4+6=12kΩ, circuit adopts form as

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Now 12kΩ and 12v are in series so we convert it into current source

I=V/R=12/12=1mA

Circuit will adopt shape as

6kΩ and 12kΩ are in parallel, their equivalent is

12x6/12+6=4kΩ

As 1mA and 4kΩ become in parallel and can be converted into voltage source

V=IXR=4x1=4v and 4kΩ become in series

Since 4v and 12v are in series and can be add up

Also 2kΩ and 4kΩ in series so 2k+4k=6kΩ, circuit can be redrawn as

Thus circuit has been transformed into a single voltage source as required

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Q.2: 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:

As we know in superposition method we consider the effect of each source one by one

and ignore other source and then add result of each source to find v0

By using super position principle to the given circuit

V =V +V .........................(1)0 01 02

Consider only 10V voltage source, We ignore effect of 4mA and open circuit it to find V01.

V01 can be determined by voltage divider, loop method or nodal analysis, we use nodal

analysis here

In the above circuits,

2 1V =V -V ..............................(2)01

Applying KCL at node V1

( )

1 1 2 1 04 2 123 6 6

1 1 2 1 012

10 6 0 ........................... 31 2

V V V V

k k kV V V V

k

V V

−+ + =

+ − +=

− =

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( )4

102 1 2 2 02 4 4

2 2 102 1 2 2 0

4 2 10 ................. 42 1

V V V V

k k kV V V V

V V

− −+ + =

− + − +=

− =

Solving equations (3) & (4) we can get,

2.143 &1

3.52

V V

V V

=

=

From ………… (2)

3.5 2.1401

1.401

V

V V

= −

∴ =

Consider only 4mA current source,

i.e., 10V is short circuited and find V02 value.

V02 can be determined by current divider, loop method or nodal analysis, we use nodal

analysis here

In the above circuit,

2 1.................(5)02

V V V= −


31 1 2 1 (4 10 ) 04 2 123 6 6 48

1 1 2 1 012

10 6 48 ............................(6)1 2

V V V V

k k kV V V V

kV V

−−+ + + × =

+ − + +=

− = −

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( )

2 1 2 2 02 4 4

2 22 1 2 2 0

4

4 2 0 ..................... 72 1

V V V V

k k kV V V V

k

V V

−+ + =

− + +=

− =

( ) ( )6 & 7

,

6.86 &1

3.432

.....................(5)

3.43 6.8602

3.4302& (1)

01 02

,

1.4 3.40

4.8 .0

Solving equations

we can get

V V

V V

From

V

V V

V V values subsitute in equation

we can get

V

V V approxi Answer

= −

= −

= − +

∴ =

= +

=

Q.3: Find the Thevenin’s equivalent circuit for the network in shaded area between terminals

a, b.

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Step 1: Removing RL To find Thevenin's equivalent voltage

Remove the load resistance RL, circuit shape will be as

Step 2: Finding voltage Vth The absence of direct connection between a and b results in network with three parallel

branches. The voltage V1 and V2 can therefore be determined using the voltage divider

rule.

1

11

1 3

1

2

2

2 1

sin

6 43272 48

6 3 9

12 86472 54

12 4 16

54 48

6

th

th

th

now the voltge across R will be

u g formula

RV V

R R

VV V V

now the voltge across R will be

VV V V

V V V

V v v

V V

=+

Ω= = =

Ω + Ω Ω

Ω= = =

Ω + Ω Ω

= −

= −

=

Vth=6v

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Step 3: Finding Rth We short circuit voltage source, it provides direct connection between c and c

/

( ) ( )

( ) ( )

1 3 2 4

1 3 13

2 4 24

1 3 2 4

13 24

6*3 182

6 3 9

12*4 483

12 4 16

2 3 5

5

th

th

th

th

th

TO FIND R

R R R R R

R R R

R R R

R R R R R

AS R AND R ARE IN SERIES

R

R

= +

Ω= = = = Ω

+ Ω

Ω= = = = Ω

+ Ω

= +

= Ω + Ω = Ω

= Ω

Rth = 5Ω

Step 4: Finding Thevenin’s equivalent circuit

The Thevenin’s equivalent will be

Voltage source Vth in series wills Rth and RL

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CIRCUIT THEORY (PHY301)MARKS: 30

Due Date: 09/06/2009

Q.1: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 thecircuit diagram of each step and also mention the units of each derived value.

Solution:Remove RLFirst Remove RL(here RL is 2kΩ )and then short the terminal to calculate the IN

Calculating IN

After removal of load resistor calculate Norton current IN

IN can be calculated by any method (loop, nodal superposition, etc) we use loop analysis

Apply KVL for loops

Here IN = I1-I2

Writing KVL FOR Loop 1:

KVL FOR Loop 2:

FOR Loop 3:

Put value of I3 in equation (1) and equation (2)

Calculating RN

Calculate RN By open current source and short the voltage source

1KΩ is series with the 1KΩ resister so

And 2KΩ is parallel with 2KΩ

Now Norton equivalent circuit is;Now we put Norton IN, RN, in parallel of RL

Apply current divider rule to calculate IRL

Q.2:In the circuit shown below there are two identical diodes having n=1 and Is=10-15 A. Findthe value to obtain an output voltage Vo=2V, if a current of 2mA is drawn away from theoutput terminal by a load, what is the change in output voltage.

Solution:

We are given V0 =2V, and diode=2 so the voltage drop across each diode is 2/2 V= 1VThus I must be

I = Is e v/nVT

Where VT= 25 mV=0.025 volts, Is = 10-15 A, n = 1, v=1v

= 10-15 e1/1*0.025

=10-15 x2.354x1017

I=235.4A

If a current of 2mA is drawn away from the terminals by means of a load, the currentthough the diodes reduces to

235.4A – 0.002A = 235.398A (2mA=0.002A)

Thus the voltage across each diode changes by∆V = nVTln(i2/i1)

=1x0.025xln(235.398A)/ 235.4A

∆V = 0.025 ln (0.999)

∆V = 0.025x-0.001SoΔV =-2.51x10-4 VThe total decrease in V0

V0 = 2 x -2.51x10-4

= 5x x10-4 v

Q.3:Below are given two diodes (a), (b) with different polarities. Describe reverse/forwardbiased condition of each with justification.

ANSWERFig (a):

The positive terminal of the battery is connected to the cathode of diode andanode of diode is connected through a resistor to neutral terminal, therefore no currentwill flow and we can say that diode does not exist and it will act as an open circuit.

Fig (b):In figure (b) the negative terminal of a battery is connected to the cathode of

diode whereas the anode is attached through a resistor to the neutral terminal and diodewill be in the forward biased condition

………. Good Luck……….


Circuit Theory (Phy301)Marks: 25

Due Date: 19/06/2009

Q.1:DC analysis circuit for ideal diode is given below

(a) Draw the shape of circuit for AC analysis(b) Find the ac source voltage and diode signal voltage if id=2A.

Solution:(a) For AC analysis we will remove DC sources which are the part of the original circuitand also remove the DC source which appear in the previous effective circuit and we alsoremove the ideal circuit so our circuit will be as

(b)Source voltage can be found as

Vs = id (R + rd)Vs=2(3+1)Vs=8v

if we carefully see the ac equation circuit, it is nothing more than a voltage divider.Hence the diode signal voltage will be

Vd = Vs rd/(rd +R)

Vd=8x1/3+1Vd=2v

Q.2:The Turn ratio of a transformer is 6:3, what type of transformer is it? Find the primarycurrent and power if secondary current is 5A.

Solution;Transformer is step down as secondary winding is less than primary windings.

As The turn ratio of the transformer is equal to the voltage ratio of the two components,so that

Also we knowThe current ratio is inverse to the voltage ratio

Since the Voltage ratio of a transformer is equal to the turn ratio therefore, the previous equation can bewritten as

We have to find I1 so

Primary current is

I1=15/6=2.5APowerRatio of primary power and secondary power is

As you are not given value of primary voltage soPp/Ps = VpIp/VsIs = Np/Ns × Ns/Np = 1Primary power will equal to secondary power.

Formatted: Font: Bold

Q.3:Determine the values of VL(pk), Vave, IL(pk) and Iave for the circuit shown in the fig.

Solution:To find VL(pk), first of all we have to find V2(pk)

V2(pk) = 18Vac/0.707

V2(pk) = 25.46VpkNow peak load voltage is

VL(pk) = V2(pk)/2 – 0.7

VL(pk) =12.023VpkAnd

Vave = 2VL(pk)/ Π=0.636 VL(pk)

Vave =7.66Vdc

Now peak load current isIL(pk) = VL(pk)/RL

=12.023 /5kIL(pk) =2.405mApk

AndIave = Vave/RL

=7.66V/5kIave =1.532mA

Assignment 6(Spring 2009) (Solution)Circuit Theory (Phy301)

Marks: 20Due Date: 26/06/2009

Q.1:For the following clipper circuit, determine the VO and sketch out put waveform ifgermanium diode is used.

Solution:For positive half cycle 5.7v is transition voltage and above transition region diode will beoffSo Vo=Vi

V0=10vOutput wave form for positive region is

-Ve region:For –ve region half cycle diode will be on, here Id=0 Vd=0.3v (As for germanium diodeforward voltage is 0.3v)Vi+Vd-V=0Vi=V-VdVi =6-0.3Vi=5.7v

Output wave for –ve region is

Complete output wave form is

Q.2:

What type of GATE is given in circuit below? Find V and I for each diode.

Answer:It is Diode AND gate

D2 is ON.The applied voltage at the cathode terminal of the diode D2 is +2V and

applied voltage at the anode terminal of the diode D2 is +10V so the potential differenceis 8V. Its mean that the anode terminal of the diode D2 is more positive as compared thecathode terminal. So it will be in forward biased condition i:e D1 is ON.D1 is OFF.

The potential difference across the diode D1 is 5V. Its mean the anodeterminal of the diode D1 is more positive as compared the cathode terminal of the diodeD1. However potential difference is less than as in case of diode D2, so diode D1 is inforward biased condition but it is not as much forward biased as in the case of the diodeD2 because we are considering the case of ideal diodes we consider that the when diodeD2 is fully forward biased and it will pass more current than diode D1.

Due to the forward biasing of diode D2 this potential difference will supply onoutput and this potential difference operate the diode D1 in reverse biased.

D1 is OFFD2 is ON

ThereforeV = +2V

By applying Ohm’s LawI = (10-2) / 2kI =4mA

Circuit Theory - Solved Assignments - Spring 2009

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