A resistance `R` draws power `P` when connected to an `AC` source. If an inductance is now placed in series with the resistance, such that the impedence of the circuit becomes `Z`, the power drawn will be

Options

A. `P sqrt((R)/(Z))`
B. `P((R)/(Z))`
C. P
D. `P((R)/(Z))^(2)`

Answer

D

Explanation

A resistor is connected to an AC source
Power `(P) = V_(rms) I_(rms) = (V_(rms)^(2))/(R)`
When an inductor is connected in series with the resistor
`P. = V_(rms) I_(rms) cos phi = (V_(rms)^(2))/(Z) * (R)/(Z)`
`rArr P. = (P*R)/(Z)* (R)/(Z) = P ((R)/(Z))^(2) " "` [`:. I_(rms) = (V_(rms))/(Z)`]
`rArr P. = P ((R)/(Z))^(2)`