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Assignment 4 (Fall 2010)

Circuit Theory (PHY301)

Marks: 30

Due Date: January 6, 2011


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Using the superposition method find Vo in the network given below. Label and draw each step where it required. Write each step of the calculation to get maximum marks and also mention the units of each derived value.














Using the source transformation method find Vo in the network given below. Label and draw each step where it required. Write each step of the calculation to get maximum marks and also mention the units of each derived value.









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.





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ircuit Theory (PHY301)

Marks: 30

Due Date: January 6, 2011






Step by Step

Using superposition to find a given output can be broken down into foursteps:

1.   Isolate a source -Select a source, and set all of the remaining sources to zero. The consequencesof "turning off" these sources are explained in Open and ClosedCircuits. In summary, turning off a voltage source results in ashort circuit, and turning off a current source results in an open circuit.(Reasoning - no current can flow through a open circuit and there can be novoltage drop across a short circuit.)

2.  Find the output from the isolated source - Once asource has been isolated, the response from the source in question can be foundusing any of the techniques we've learnedthus far.

3.  Repeat steps 1 and 2for each source - Continue to choose a source,set the remaining sources to zero, and find the response. Repeat this procedureuntil every source has been accounted for.

4.  Sum the Outputs -Once the output due to each source has been found, add themtogether to find the total response.


Solution :

V =V +V .........................(1)

Applying KCL at node V1

V V −V V 1 + 1 2 + 1 =0 4k 2k 12k 3V + 6V − 6V + V1 1 2 1 =0 12k 10V − 6V = 0 ...........................( 3) S

V =V2 -V1

In the circuits,


Consider only 10Vvoltage source, We ignore effect of 4mA and open circuit it to find V01.V01 can be determined by voltage divider, loop method ornodal analysis, we use nodal analysis here

Solution: As we knowin superposition method weconsider the effect of each source one by one and ignore othersource and then add result of each source to find v0By using superposition principle to the given circuit

Applying KCL at node V2

4V − 2V = 2

.................( 4 )

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

V = 2.143V & 1 V = 3.5V 2
From ………… (2)

In the above circuit,

Applying KCL at node V1

V V −V V 1 + 1 2 + 1 + (4 ×10−3) = 0 4k 2k 12k 3V+ 6V − 6V + V + 48 1 1 2 1 =0 12k 10V − 6V = −48............................(6)

V = V2 −V1.................(5)

Consider only 4mAcurrent source, i.e., 10V is short circuited and find V02 value. V02 can be determined by currentdivider, loop method or nodal analysis, we use nodalanalysis here

V = 3.5 − 2.14 01 V = 1.4V 01

V −V V −10 V 2 1+ 2 + 2 =0 2k 4k 4k 2V − 2V + V−10 + V 2 1 2 2 =0

Applying KCL at node V2

V= 4.8V approxi.


Solution :


by source transformation

Now2mA source is in parallel with 3k resistor.So itcan be changed to a voltage source of value = 2m x 3k (by ohm’s Law)

=6 Volts.

3kresistor will become in series withthis source as shown in the circuit below

Nowthe combined effect of these two source will be 9 volts.


Vth= 9 volts

Third step:

Calculating Rth
Rth =3k

Fourth step:

Calculatingthe unknown quantity.

Aftercalculating Vth and Rth, re-inserting the load resistance RL in thecircuit in series with Rth andconsidering the Vth as a battery in series withthese two resistances.

V0 = (6k/9k) x9
V0= 6volts


Solution :

Removing load resistance RL= 1KΩ

In the above circuit Ix= 0 because current is not completing its path due to open circuit so, value ofdependent current source is also zero.

VTH = 6 V Now for RTHwe will shortthe open terminal of the Thévenin’s circuit andcalculate the Isc and then divide VTH with Isc to calculate RTH

Applying KCL at node VSum of all the currents leaving the junction = sum of all the currents enteringthe junction V V+6 + = 2Ix 1k 2k Here V+6 Ix = 2k V + 1k 2k V+6 =2 2k V+6

V + 1k V + 1k

V+6 = 2k V+6 2k

V+6 1k V+6 =0 1k =0

2V + V + 6 – 2[V + 6] 2k 2V + V + 6 – 2[V + 6] = 02V + V + 6 – 2V - 12 = 0 V=6V Now 6+6 Ix = 2k

Ix = 6 mA = ISC RTH = VTH ISC 6V RTH = 6 mA

RTH = 1 kΩ


According to Voltage divider rule: 1k V0 = ×6V1k+1k

V0 = 3 V


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