Question

A resistor of resistance 'R' draws power 'P' when connected to an AC source. If an inductance is now placed in series with R, such that the impedance of the circuit becomes 'Z', the power drawn will be

• A. \( P\left(\frac{R}{Z}\right) \)
• B. \( P\left(\frac{R}{Z}\right)^3 \)
• C. \( P\left(\frac{R}{Z}\right)^2 \)
• D. \( P\sqrt{\frac{Z}{R}} \)

Hint:

Alternatively, you can write the new power as \( P' = \frac{V_{\text{rms}}^2}{Z} \cos\phi \), where \( \cos\phi = \frac{R}{Z} \) is the power factor. This gives \( P' = \frac{V_{\text{rms}}^2 R}{Z^2} \). Since \( V_{\text{rms}}^2 = P R \), the expression simplifies immediately to \( P' = P \left(\frac{R}{Z}\right)^2 \).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Initially, a pure resistor \( R \) is connected across an AC source and draws power \( P \). Then, an inductor is connected in series, increasing the circuit's total impedance to \( Z \). We need to express the new power drawn \( P' \) in terms of the initial power \( P \), resistance \( R \), and impedance \( Z \).

Step 2: Key Formula or Approach:
1. Initial Power (purely resistive circuit):
\[ P = \frac{V_{\text{rms}}^2}{R} \implies V_{\text{rms}}^2 = P R \]
2. Power in an L-R Series AC Circuit:
\[ P' = I_{\text{rms}}^2 R \]
where \( I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} \).

Step 3: Detailed Explanation:
Substitute \( I_{\text{rms}} \) into the formula for \( P' \):
\[ P' = \left(\frac{V_{\text{rms}}}{Z}\right)^2 R = \frac{V_{\text{rms}}^2 R}{Z^2} \]
Using the relationship \( V_{\text{rms}}^2 = P R \) obtained from the first case:
\[ P' = \frac{(P R) R}{Z^2} = P \frac{R^2}{Z^2} = P \left(\frac{R}{Z}\right)^2 \]

Step 4: Final Answer:
The power drawn in the modified circuit is \( P\left(\frac{R}{Z}\right)^2 \), which corresponds to option (C).