Question

For LCR circuit shown in the figure, value of angular frequency at which power of the circuit becomes half of the maximum power may be

L = 12.5 H C = 5 mF R = 75 $\Omega$

Answer 1

2 rad/s or 8 rad/s

Answer 2

The power dissipated in an LCR circuit is given by: $P = I_{rms}^2 R = \frac{V_{rms}^2}{Z^2} R$ where $Z = \sqrt{R^2 + (X_L - X_C)^2}$ is the impedance, $X_L = \omega L$ is the inductive reactance, and $X_C = \frac{1}{\omega C}$ is the capacitive reactance.

The maximum power ($P_{max}$) occurs at resonance, where $X_L = X_C$. At resonance, the impedance is $Z_{min} = R$. So, $P_{max} = \frac{V_{rms}^2}{R}$.

We are looking for the angular frequencies ($\omega$) at which the power becomes half of the maximum power, i.e., $P = \frac{1}{2} P_{max}$. $\frac{V_{rms}^2 R}{R^2 + (X_L - X_C)^2} = \frac{1}{2} \frac{V_{rms}^2}{R}$

Simplifying the equation: $\frac{R}{R^2 + (X_L - X_C)^2} = \frac{1}{2R}$ $2R^2 = R^2 + (X_L - X_C)^2$ $R^2 = (X_L - X_C)^2$

Taking the square root of both sides: $X_L - X_C = \pm R$ Substituting the expressions for $X_L$ and $X_C$: $\omega L - \frac{1}{\omega C} = \pm R$

This leads to two quadratic equations for $\omega$:

Case 1: $\omega L - \frac{1}{\omega C} = R$ Multiplying by $\omega C$: $\omega^2 LC - 1 = R \omega C$ $\omega^2 LC - R C \omega - 1 = 0$

Case 2: $\omega L - \frac{1}{\omega C} = -R$ Multiplying by $\omega C$: $\omega^2 LC - 1 = -R \omega C$ $\omega^2 LC + R C \omega - 1 = 0$

Let's calculate the resonant angular frequency $\omega_0$ and the term $\frac{R}{2L}$. Given: $L = 12.5 \text{ H}$, $C = 5 \text{ mF} = 5 \times 10^{-3} \text{ F}$, $R = 75 \Omega$.

Resonant angular frequency: $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{12.5 \times 5 \times 10^{-3}}} = \frac{1}{\sqrt{62.5 \times 10^{-3}}} = \frac{1}{\sqrt{0.0625}} = \frac{1}{0.25} = 4 \text{ rad/s}$

Calculate $\frac{R}{2L}$: $\frac{R}{2L} = \frac{75}{2 \times 12.5} = \frac{75}{25} = 3 \text{ rad/s}$

The solutions to the quadratic equations $\omega^2 LC \mp R C \omega - 1 = 0$ are given by the formula: $\omega = \pm \frac{R}{2L} + \sqrt{\omega_0^2 + \left(\frac{R}{2L}\right)^2}$ (taking only positive frequencies)

For Case 1 ($\omega L - \frac{1}{\omega C} = R$), the higher half-power frequency $\omega_2$ is: $\omega_2 = \frac{R}{2L} + \sqrt{\omega_0^2 + \left(\frac{R}{2L}\right)^2}$ $\omega_2 = 3 + \sqrt{4^2 + 3^2} = 3 + \sqrt{16 + 9} = 3 + \sqrt{25} = 3 + 5 = 8 \text{ rad/s}$

For Case 2 ($\omega L - \frac{1}{\omega C} = -R$), the lower half-power frequency $\omega_1$ is: $\omega_1 = -\frac{R}{2L} + \sqrt{\omega_0^2 + \left(\frac{R}{2L}\right)^2}$ $\omega_1 = -3 + \sqrt{4^2 + 3^2} = -3 + \sqrt{16 + 9} = -3 + \sqrt{25} = -3 + 5 = 2 \text{ rad/s}$

Thus, the two angular frequencies at which the power of the circuit becomes half of the maximum power are 2 rad/s and 8 rad/s.