Answer
(a)A,B,C
Work Step by Step
we have
$P=VI$
Doubling the voltage and reducing the current by a factor of two
$ P' = 2V* \frac{1}{2}I =P$
Also $P = \frac{V^2}{R}$
Doubling the voltage and increasing the resistance by a factor of four
$ P' = \frac{(2V)^2} {4R} = \frac {V^2}{R} =P$
Also $ P = I^2R$
Doubling the current and reducing the resistance by a factor of four
$ P' = (2I)^2 (\frac{1}{4}R) = I^2R =P$
