Question

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

• A. \(\left(\frac{R}{Z}\right)^{2}\)
• B. \(\sqrt{\frac{R}{Z}}\)
• C. \(\frac{R}{Z}\)
• D. \(1\)

Hint:

Only the resistor dissipates power, so compare V squared over R with the current squared times R in the series circuit.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Let the r.m.s. source voltage be \(V\). With only the resistor, \(P=\frac{V^{2}}{R}\).

Step 2: Series circuit
With the inductor in series, the impedance is \(Z\) and the current is \(I=\frac{V}{Z}\). Only the resistor dissipates power: \(P_{1}=I^{2}R=\frac{V^{2}R}{Z^{2}}\).

Step 3: Ratio
\(\frac{P_{1}}{P}=\frac{V^{2}R/Z^{2}}{V^{2}/R}=\frac{R^{2}}{Z^{2}}=\left(\frac{R}{Z}\right)^{2}\).

Final Answer:
The ratio of powers is (R over Z) squared. \[ \boxed{\text{(A) }\left(\dfrac{R}{Z}\right)^{2}} \]