Resonant Frequency in LCR Circuit – Rankers Physics

LR, RC and LCR Circuits: Practice Problem & Solution

The value of capacitance \(C\) in an LCR series circuit with inductance \(L = 50\) mH, connected with a voltage source of angular frequency \(100\) rad/s so that current in the circuit becomes maximum, is:
1 mF
2 mF
1 \(\mu\)F
4 \(\mu\)F

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of LR, RC and LCR Circuits. Understanding the underlying formula is key to arriving at the correct answer below:

Current is maximum at resonance, where \(\omega = \frac{1}{\sqrt{LC}}\). Solving for \(C\): \(C = \frac{1}{\omega^2 L} = \frac{1}{(100)^2 \times 50 \times 10^{-3}} = 2 \times 10^{-3}\)⇒ C = 2 mF.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Assertion (A): Choke coil is preferred over a resistor to adjust current in an \( \text{ac} \) circuit. Reason (R): Power factor for inductance is zero.  
2. For a series RLC circuit \(R = X_L = 2X_C\). The impedance of the circuit and phase difference between \(V\) and \(i\) will be:
3. An ac current flowing in a circuit is given by \(i = i_0 sin \omega t\). The minimum time taken to reach zero to rms value of current is
4. The phase difference between voltage and current in series LCR circuit at half power frequency is:
5. An L-C-R series circuit with \(100\ \Omega\) resistance is connected to an ac source of \(100\text{ V}\) and angular frequency \(300\text{ rad/s}\). When the capacitance is removed, the current lags behind the voltage by \(45^\circ\). When the inductance is removed, the current leads the voltage by \(45^\circ\). The current flowing in the original circuit will be: