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CBSE Class 12 Physics 2017 Outside Delhi Set 1 Paper

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Question : 11
Total: 26
SECTION-C

(a) The potential difference applied across a given resistor is altered so that the heat produced per second increases by a factor of 9. By what factor does the applied potential difference change?
(b) In the figure shown, an ammeter A and a resistor of 4Ω are connected to the terminals of the source. The emf of the source is 12V having an internal resistance of 2 Ω. Calculate the voltmeter and ammeter readings.
Solution:  
(a) The factor by which the potential difference changes
(b) Voltmeter reading
Ammeter Reading
(a) (H=‌
V2
R
)
∴V increases by a factor of √9=3
(b) Ammeter Reading I‌=‌
V
R+r
‌=‌
12
4+2
A=2A
Voltmeter Reading V=E−Ir
=[12−(2×2)]V=8V
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