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Physics Question on Electromagnetic induction

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

A

P(RZ)P\big(\frac{R}{Z}\big)

B

PP

C

P(RZ)2P\big(\frac{R}{Z}\big)^2

D

PRZP \sqrt{\frac{R}{Z}}

Answer

P(RZ)2P\big(\frac{R}{Z}\big)^2

Explanation

Solution

Case I : P=VrmsIrmsP=V_{rms}I_{rms}
=Vrms×VrmsR=V_{rms}\times\frac{V_{rms}}{R}
P=Vrms2R⇒Vrms2=PRP=\frac{V^2_{rms}}{R}\Rightarrow V^2_{rms}=PR
Case II : Power drawn in LR circuit
P′=VrmsIrmscosϕ=Vrms×VrmsZ×RZP'=V_{rms}I_{rms}cos\phi=V_{rms}\times\frac{V_{rms}}{Z}\times\frac{R}{Z}
=Vrms2RZ2=PR×RZ2=V^2_{rms}\frac{R}{Z^2}=PR\times\frac{R}{Z^2}
\hspace20mm [Using eqn (i)]
P′=p(RZ)2P'=p\left(\frac{R}{Z}\right)^{2}