Question

Question #ee8e8

Answer 1

(A)

Explanation:

From Ohm's law we know that
`V=IR` ......(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
`2V=I_dR` ......(3)
Dividing (3) with (1)
`(2V)/V=(I_dR)/(IR)`
`=>I_d=2I`

Similarly from (2) we get
`P_d=(2V)^2/R` .....(4)
Dividing (4) with (2) we get
`P_d/P=((2V)^2/R)/(V^2/R)`
`=>P_d=4P`