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CBSE Class 12 Physics 2017 Outside Delhi Set 1 Paper

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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 AA and a resistor of 4Ω4 \Omega are connected to the terminals of the source. The emf of the source is 12V12 V having an internal resistance of 2 Ω\Omega. Calculate the voltmeter and ammeter readings.
Solution:  
(a) The factor by which the potential difference changes
(b) Voltmeter reading
Ammeter Reading
(a) (H=;V2R)\left(H=\\;\frac{V^2}{R}\right)
∴V\therefore V increases by a factor of 9=3\sqrt{9}=3
(b) Ammeter Reading I;=;VR+rI\\;=\\;\frac{V}{R+r}
;=;124+2A=2A\\;=\\;\frac{12}{4+2} A=2 A
Voltmeter Reading V=E−IrV=E-I r
=[12−(2×2)]V=8V=[12-(2 \times 2)] V=8 V
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