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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 38Ω 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 Ω 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,
200−38i−10=0
∴‌‌i=‌
190
38
A=5A

OR
Stating the formula
Calculating r
We have, r=(‌
l1
l2
−1)R=(‌
l1−l2
l2
)R
∴‌‌r‌=(‌
350−300
300
)×9Ω
‌=‌
50
300
×9Ω=1.5Ω
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