Question #ee8e8

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Question #ee8e8

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Answer 1 · 2017-02-19T10:15:30

(A)

Explanation:

From Ohm's law we know that
$V = I R$ $V=IR$ ......(1)
Also we know that Power $P$ $P$ dissipated in a circuit is
$P = \frac{V^{2}}{R}$ $P=V^2/R$ ........(2)

When Voltage $V$ $V$ is doubled
from (1) we get
$2 V = I_{d} R$ $2V=I_dR$ ......(3)
Dividing (3) with (1)
$\frac{2 V}{V} = \frac{I_{d} R}{I R}$ $(2V)/V=(I_dR)/(IR)$
$\Rightarrow I_{d} = 2 I$ $=>I_d=2I$

Similarly from (2) we get
$P_{d} = \frac{{(2 V)}^{2}}{R}$ $P_d=(2V)^2/R$ .....(4)
Dividing (4) with (2) we get
$\frac{P_{d}}{P} = \frac{\frac{{(2 V)}^{2}}{R}}{\frac{V^{2}}{R}}$ $P_d/P=((2V)^2/R)/(V^2/R)$
$\Rightarrow P_{d} = 4 P$ $=>P_d=4P$

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Question #ee8e8

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