CBSE Class 12 Physics 2017 Outside Delhi Set 1 Paper

Question : 11 of 26

Marks: +1, -0

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

4 Ω

are connected to the terminals of the source. The emf of the source is

12 {\V}

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=\;{V^2}/{R})

∴ V

increases by a factor of

{√{9}}=3

(b) Ammeter Reading

I\;=\;{V}/{R+r}

\;=\;{12}/{4+2} {\A}=2 {\A}

Voltmeter Reading

V=E-I r

=[12-(2 × 2)] {\V}=8 {\V}

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