Question

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. $P\big(\frac{R}{Z}\big)$
• B. $P$
• C. $P\big(\frac{R}{Z}\big)^2$
• D. $P \sqrt{\frac{R}{Z}}$

Answer 1

The Correct Option is C

To find the power drawn by the circuit when an inductance is placed in series with a resistance, and the impedance of the circuit becomes \( Z \), we need to understand the relationship between power, resistance, and impedance in AC circuits.

1. Initially, the resistance \( R \) draws power \( P \) from the AC source. The power formula for a purely resistive AC circuit is given by: P = I^2 R, where \( I \) is the current.
2. When an inductance is added in series, the impedance \( Z \) of the circuit is given by \( Z = \sqrt{R^2 + (X_L)^2} \), where \( X_L \) is the inductive reactance.
3. The current through the circuit when the impedance is \( Z \) can be expressed as: I' = \frac{V}{Z}, where \( V \) is the source voltage.
4. With this current, the power drawn by the circuit with impedance \( Z \) becomes: P' = (I')^2 R = \left(\frac{V}{Z}\right)^2 R = \frac{V^2 R}{Z^2}.
5. From the original power equation, \( P = \frac{V^2}{R} \), we can substitute to express the new power in terms of the initial power: P' = P \left(\frac{R}{Z}\right)^2.

Therefore, the correct answer is P\left(\frac{R}{Z}\right)^2. This shows that the power drawn reduces when an inductance is added, as the impedance \( Z \) is greater than the resistance \( R \), making the term \( \frac{R}{Z} \) less than 1.