Q181 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Physics
A resistance 'R' draws a power 'P' when connected to A.C. source. If an inductor is now placed in series with the resistance, such that the impedance of the circuit becomes 'Z', the power drawn is $'P_1'$. The ratio of $P_1 / P$ is
- Correct answerA$\left(\dfrac{R}{Z}\right)^2$
- B$\sqrt{\dfrac{R}{Z}}$
- C$\dfrac{R}{Z}$
- D$1$
Solution
Let the RMS voltage of the AC source be $V$.
When only resistance $R$ is connected, the power drawn is $P = \dfrac{V^2}{R}$.
When an inductor is connected in series, the impedance becomes $Z$. The RMS current in the circuit is $I = \dfrac{V}{Z}$.
The power drawn in the L-R circuit is $P_1 = I^2 R = \left(\dfrac{V}{Z}\right)^2 R = \dfrac{V^2 R}{Z^2}$.
The ratio of the powers is $\dfrac{P_1}{P} = \dfrac{\dfrac{V^2 R}{Z^2}}{\dfrac{V^2}{R}} = \dfrac{R^2}{Z^2} = \left(\dfrac{R}{Z}\right)^2$.
Answer: $\left(\dfrac{R}{Z}\right)^2$
When only resistance $R$ is connected, the power drawn is $P = \dfrac{V^2}{R}$.
When an inductor is connected in series, the impedance becomes $Z$. The RMS current in the circuit is $I = \dfrac{V}{Z}$.
The power drawn in the L-R circuit is $P_1 = I^2 R = \left(\dfrac{V}{Z}\right)^2 R = \dfrac{V^2 R}{Z^2}$.
The ratio of the powers is $\dfrac{P_1}{P} = \dfrac{\dfrac{V^2 R}{Z^2}}{\dfrac{V^2}{R}} = \dfrac{R^2}{Z^2} = \left(\dfrac{R}{Z}\right)^2$.
Answer: $\left(\dfrac{R}{Z}\right)^2$
Hence, the correct answer is option: 'A'.