Question

In LCR series circuit if the frequency is increased, the impedance of the circuit

• A. increases
• B. decreases
• C. either increases or decreases
• D. first decreases then become minimum and then increases.

Answer 1

Answer: (D)

first decreases then becomes minimum and then increases.

Answer 2

For a series LCR circuit, the impedance is given by:

$$Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$$

At very low frequency ($\omega$ is small), $\frac{1}{\omega C}$ is very high compared to $\omega L$, so $|\omega L - \frac{1}{\omega C}|$ is large and hence $Z$ is high. As the frequency increases, $\frac{1}{\omega C}$ decreases while $\omega L$ increases. The difference $|\omega L - \frac{1}{\omega C}|$ reduces, and the impedance decreases until resonance (where $\omega L = \frac{1}{\omega C}$ and $Z = R$; minimum value). Beyond resonance, the inductive reactance dominates, so the impedance increases.

Thus, as the frequency is increased from a very low value, the impedance first decreases (tending to a minimum at resonance) and then increases.

Minimal Explanation:

The impedance in a series LCR circuit is high at low frequency due to a high capacitive reactance, decreases to a minimum at resonance, and then increases as the inductive reactance dominates.