Question #ee8e8

Feb 19, 2017

(A)

Explanation:

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

When Voltage $V$ is doubled
from (1) we get
$2 V = {I}_{\mathrm{dR}}$ ......(3)
Dividing (3) with (1)
$\frac{2 V}{V} = \frac{{I}_{\mathrm{dR}}}{I R}$
$\implies {I}_{d} = 2 I$

Similarly from (2) we get
${P}_{d} = {\left(2 V\right)}^{2} / R$ .....(4)
Dividing (4) with (2) we get
${P}_{d} / P = \frac{{\left(2 V\right)}^{2} / R}{{V}^{2} / R}$
$\implies {P}_{d} = 4 P$