# Fall 2011_PHY301_1_SOL

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```					Assignment 1(Fall 2011)
(Solution)
Circuit Theory (Phy301)
Marks: 25
th
Due Date: 27 October, 2011

Q.1
Find the equivalent resistance at point a, b of given circuit. Write each step of the
calculation to get maximum marks. Draw the circuit diagram of each step otherwise you

Solution:
Step 1:
2Ω resistor is in series with 2Ω resistor
So 2Ω + 2Ω = 4Ω

Step 2:
4Ω resistor is in parallel with 4Ω resistor
So      1/R = 1/R + 1/R
eq        1       2
= 1/4Ω + 1/4Ω
= (4 × 4)/(4 + 4)
= 16/8
= 2Ω
Step 3:
6Ω Resistor is in series with 2Ω Resistor
So 6Ω + 2Ω = 8Ω

Step 4:
8Ω resistor is in parallel with 8Ω resistor
So       1/Req = 1/R1 + 1/R2
= 1/8Ω + 1/8Ω
= (8 × 8)/(8 + 8)
= 64/16
= 4Ω

Step 5:
5Ω resistor is in series with 4Ω resistor
So 5Ω + 4Ω = 9Ω
Q.2
Using the division rule, find V and I mentioned in the given circuit.

Sol:

Finding Current I:
As 10Ω resistor is in series with current source therefore current remains same
Current will divide at point Node. So current I through 4Ω can be found using
current divider formula.
R2              2
I          *12 A        *12
R2  R3         24
2      24
 *12 
6      6
I  4A
So current flow through 4Ω Resistor will be 4A.
Now current flow through 2Ω resistor is
R3             4
I2           *12 A      *12
R3  R 2         42
4     48
 *12      8A
6     6
I 2  8A
Current flow through 2Ω resistor is 8A.

Finding Voltage V:
As we know that
V = IR
So we can find voltage across 2Ω resistor as
V = I2 R
=8×2
= 16 V
Voltage across 2Ω resistor is 16V.
Q.3
For the given circuit below, find current flow through each element, for the case when
(a) Point a, b is open.
(b) Point a, b is closed.

Solution:
(a) When Point a, b is open
As we know that P = V × I
So       I = P/V
= 4/20
= 0.2A
So current I1 is 0.2A
As point a, b is open therefore current I2 and I3 will be zero
(b) When point a, b is closed
First we see that all elements are in parallel, so same 20v voltage drops across
each element.
As we know that P = V × I
So        I = P/V
= 4/20
= 0.2A
So current I1 is 0.2A
Finding I2
As    V = IR
I = V/R
I2 = 20/5000
I2 = 0.004A
So current I2 is 0.004A=4mA
Finding I3
As P=VI
I = P/V
I3 = 10/20
I3 = 0.5A
So current I3 is 0.5A

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Description: Past assignments, CS201, CS301, CS304, CS601, Pak301, Eng201, PHY301