Question:

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 impedance of the circuit becomes 'Z', the power drawn will be?

P(RZ)2

P√RZ

P(RZ)

Power P = irms2R = (Vrms/R)2R
Power when inductor is put in series with R:
P' = (Vrms/Z)2R
P'/P = R/Z2 R2 = (R/Z)2
P' = P(R/Z)2