CBSE Class 12 Physics 2018 Delhi Set 1 Paper

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Question : 7
Total: 26
A 10V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 as shown in the figure. Find the value of current in the circuit.

OR
In a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350cm. When a resistance of 9 is used in the external circuit of the cell, the balance point shifts to 300cm. Determine the internal resistance of the cell.
Solution:  
Writing the equation
Finding the current
By Kirchoff's law, we have, for the loop ABDC,
20038i10=0
i=
190
38
A
=5A


OR
Stating the formula
Calculating r
We have, r=(
l1
l2
1
)
R
=(
l1l2
l2
)
R

r=(
350300
300
)
×9

=
50
300
×9
=1.5
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