phy301 solution read

On Thu, Oct 27, 2011 at 9:46 AM, INJURT HEART <aqeelcheeku@gmail.com> wrote:

Phy301 Assignment No. 1 Solution

Wednesday, October 19, 2011 Posted In PHY Edit This

Assignment 1(Fall 2011)

Circuit Theory (Phy301)

Marks: 25

Due Date: 20th October, 2011

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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 will lose your marks

Q.2

Using the division rule, find V and I mentioned in the given circuit.

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:

Download from below link:


Phy301 assignment solution


another:
Phy301 assignment solution

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 will lose your marks.

1) On the right side 2 ohm and 2 ohm resistors are in series so total resistance = 2 + 2 = 4ohms

Replace the two resistors with a single 4 ohm resistor

2) Now 4 ohm and 4 ohm resistor are in series so total resistance = 4 + 4 =8 ohms 

Replace the two resistors with a single 8 ohm resistor

3) Now 8 ohm resistor is in series with 6 ohm resistor, so TR = 8 + 6 = 14 ohms

Replace the two resistors with a single 14 ohm resistor

4) Now 14 ohm resistor is in parallel with 8 ohm resistor so TR = 14.8/14+8 = 112/22

= 5 ohms approx.

Replace the two resistors with a single 5 ohm resistor

5) Now the 5 ohm and 5ohm resistor are in series so TR = 5 + 5 = 10 ohms

Q.2

Using the division rule, find V and I mentioned in the given circuit.

Using division rule of current:

1) Current I along 2 ohm resistor is (4/4+2) x 12A = (4/6) x 12 = 0.666 x 12= 8 A approx

so V = IR = 8x 2 = 16V

2) Current along 4 ohm resistor is (2/4+2) x 12 = (2/6) x 12 = 4 A

so I = 4A

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.

Please try to solve this by yourself. It is very easy.

Hint when point a,b is open I2 and I3 = 0

use the following formulae:

Power (watt) = V x I

Current divider rule as in Q2.

Also draw the circuits yourselves.


Another Solution:



QUESTION.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 will lose your marks.


ANSWER.1
On the right side 2 ohm and 2 ohm resistors are in series so,
Total resistance = 2 + 2 = 4


Now, 4 ohm and 4 ohm resistor are in parallel so,
Total resistance = (4 x 4) /4 + 4 = 2 ohms 


2 ohm resistor is in series with 6 ohm resistor so,
Total resistance = 2 + 6 = 8 ohms 


Now, 8 ohm resistor is in parallel with 8 ohm resistor so,
Total resistance = (8x8) / (8+8)
= 4 ohms 


The 5 ohm and 4ohm resistor are in series so,
Total resistance = 5 + 4 = 9 ohms 


__________________________________________
QUESTION.2
Using the division rule, find V and I mentioned in the given circuit.






ANSWER.2
Using division rule of current: 


Voltage across 2 ohm resistor is:
V = IR = 8 x 2 = 16v


Current along 4 ohm resistor is:
(2/ (4+2)) x 12 = (2/6) x 12 = 4A 


_____________________________________________
QUESTION.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.


ANSWER.3


a) Point a, b is open. 


When point a, b is open then,


I2 and I3 = 0


Current I1 is:
Power (watt) = V x I 
I=P/V
I=4/20
I=0.2A




b) Point a, b is closed.


When point a, b is close then,


Current I2 is:
V=IR
I = V/ R = 20 / 5000 = 4 mA


Current I1 is:
Power (watt) = V x I 
I=P/V
I=4/20
I=0.2A


Current I3 is:
Power (watt) = V x I 
I=P/V
I=10/20
I=0.5A

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