JEE Physics/ Lakshmikanta Satapathy/ Direct Current QA part 13/ Question on finding the potential difference between two points in a series parallel network of resistors connected to a DC source

1. Physics Helpline
L K Satapathy
Direct Current - 13
Distribution of Current
1 3
3 1
A
B
10V

2. Physics Helpline
L K Satapathy
Direct Current - 13
Cells in Parallel
Answer :
Question : A cell of emf 10 V and negligible internal
resistance is connected to four resistors as shown in
the figure. Then the potential difference ( VA – VB ) is
(a) – 2 V (b) + 2 V (c) – 5 V (d) + 5 V
1 3
3 1
A
B
10V
1 3
3 1
A
B
C D
10V
Resistance of branch CAD = 1 + 3 = 4
Resistance of branch CBD = 3 + 1 = 4
 Circuit resistance = 4 and 4 in parallel = 2
Also , Circuit voltage = 10V
 Circuit current I = 5 A
Current is equally shared by the two branches
2.5CAD CBDI I A  
4
4
10V

3. Physics Helpline
L K Satapathy
Direct Current - 13
Cells in Parallel
1 2.5 2.5 . . . (1)C A CA CAV V R I V      
(2) (1) 7.5 2.5 [5 ]A B AnsV V V       
Correct option = (d)
We are required to find ( VA – VB )
In the branch CAD , Current flows from C to A
 C is at a higher potential than A
3 2.5 7.5 . . . (2)C B CB CBV V R I V      
In the branch CBD , Current flows from C to B
 C is at a higher potential than B
1 3
3 1
A
B
C D
10V

4. Physics Helpline
L K Satapathy
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