NEET 2015 Physics AC Circuits MCQ Question

🧠 NEET Explanation

Given: A resistance $R$ draws power $P$ when connected to an AC source.

For a pure resistance,

$$P=\frac{V_{\rm rms}^{2}}{R}$$

So,

$$V_{\rm rms}^{2}=PR$$

Now an inductance is connected in series. The new impedance is $Z$.

The current becomes:

$$I=\frac{V_{\rm rms}}{Z}$$

Only the resistance $R$ consumes average power; the inductor does not consume average power.

Therefore,

$$P' = I^2R$$ $$P'=\left(\frac{V_{\rm rms}}{Z}\right)^2R$$

Using $V_{\rm rms}^2=PR$:

$$P'=\frac{PR^2}{Z^2}$$ $$\boxed{P'=P\left(\frac{R}{Z}\right)^2}$$

✅ Answer: Option (1)

⚡ Quick Revision

Pure R:

$$P=\frac{V^2}{R}$$

R + L series:

$$P=I^2R$$ $$\boxed{P'=P\left(\frac{R}{Z}\right)^2}$$

👉 NEET Tip: An ideal inductor has zero average power consumption; only $R$ dissipates power.