Question

With gradual increase in frequency of an a.c. supply, the impedance of an LCR series circuit

• A. increases
• B. decreases
• C. remains constant
• D. first decreases, becomes minimum and then increases

Hint:

Impedance falls to its minimum at resonance and then rises.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For a series LCR circuit, \(Z = \sqrt{R^2 + \left(\omega L - \dfrac1{\omega C}\right)^2}\).

Step 2: Low frequency:
At low frequency \(\dfrac1{\omega C}\) is large, so the reactance is large and capacitive. \(Z\) is large.

Step 3: At resonance:
As \(\omega\) increases, \(\dfrac1{\omega C}\) falls and \(\omega L\) rises until \(\omega L = \dfrac1{\omega C}\), where \(Z = R\), the minimum value.

Step 4: Beyond resonance:
For higher \(\omega\), \(\omega L\) dominates and \(Z\) rises again.
So the impedance first decreases, becomes minimum and then increases, option (D). The other options describe only one part of this behaviour.

Final Answer:
Z dips to R at resonance, then increases again. \[ \boxed{\text{(D) }\text{first decreases, becomes minimum and then increases}} \]