JEE Physics/ Lakshmikanta Satapathy/ Direct current QA part 11/ JEE question on Kirchhoff's loop rule for networks solved with the related concepts

1. Physics Helpline
L K Satapathy
Direct Current - 11
Kirchhoff’s Loop Rule
60
R
A
4V 1V

2. Physics Helpline
L K Satapathy
Direct Current - 11
Cells in Parallel
Answer :
Question : In the circuit shown in the figure, the
reading of the ammeter is zero. If the resistance of
the ammeter and internal resistance of the cells are
negligible, then the value of R is
(a) 10  (b) 20  (c) 30  (d) 40 
60
R
A
4V 1V
60
R
A CB
D
4V 1V
I
I
Let the current through branch AB = I
It is given that ammeter (A) reads zero.
 Current in the branch BC = 0
 Current in the branch BD = I

3. Physics Helpline
L K Satapathy
Direct Current - 11
Cells in Parallel
In the loop ABDA
( ) ( ) ( ) 0 . . . (1)A B B D D AV V V V V V     
60
R
A CB
D
4V 1V
I
I
Current flows from A to B ( )A BV V 
60A BV V I  
Current flows from B to D ( )B DV V 
B DV V RI  
A is connected to the +ve and D to –ve of the cell ( )A DV V 
4D AV V volts   
(1) 60 ( 4) 0I RI      (60 ) 4I R  
4
. . . (2)
60
I
R
 


4. Physics Helpline
L K Satapathy
Direct Current - 11
Cells in Parallel
60
R
A CB
D
4V 1V
I
I( ) ( ) ( ) 0 . . . (3)B C C D D BV V V V V V     
(3) 0 1 ( ) 0 1RI RI       
1
. . . (4)I
R
 
In the loop BCDB
Current from B to C = 0 0B CV V  
C is connected to the +ve and D to –ve of the cell ( )C DV V 
1C DV V volt  
B D D BAlso V V RI V V RI     

5. Physics Helpline
L K Satapathy
Direct Current - 11
Cells in Parallel
Correct option = (b)
3 60R 
4 1
(2) (4) 4 60
60
and R R
R R
    

0 [ ]2 nR A s  

6. Physics Helpline
L K Satapathy
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