Question

3. Given
E1 = 3 V, E2 = 2 V, E = I V, R=n=r2=r3 = 1 ohm
Find the potential difference between A and B.
​

Answer 1

Answer:

Explanation:

(a) Applying Kirchoff's loop law to mesh PLMQP and PLMQONP in the figure shown below, we have

i  

​  

−i  

1

​  

r  

1

​  

=2volt

(b) The figure shows the circuit when point A is connected to point B and r  

2

​  

 is short-circuited.

Applying Kirchoff's junction rule at P, we get

i=i  

1

​  

+i  

2

​  

+i  

3

​  

....(iv)

Applying Kirchoff's law to mesh ABMLA

i  

1

​  

r  

1

​  

=E  

1

​  

−E  

2

​  

;i  

1

​  

=1amp

Applying Kirchoff's law to mesh ANOQML

i  

1

​  

r  

1

​  

+i  

3

​  

r  

3

​  

=E  

1

​  

−E  

3

​  

ori  

1

​  

−i  

3

​  

=2....(v)

From above equations

i  

1

​  

=1amp,

i  

2

​  

=2amp

i  

3

​  

=−1amp (direction of current is opposite)

So, current through resistor R will be I=I  

1

​  

+I  

2

​  

+I  

3

​  

=2amp