Question

A series LCR circuit containing a resistance of 120 $\Omega$ has angular resonance frequency 4 × $10^5$ rad/sec. At resonance the voltages across resistance and inductance are 60 V and 40 V respectively, then choose the correct option/s.

• A. The value of L = 2 × $10^{-4}$ H
• B. The value of C = $\frac{1}{32}$ × $10^{-6}$ F
• C. At angular frequency 8 × $10^5$ rad/s the current will lag by $30^\circ$ w.r.t. voltage
• D. At angular frequency 8 × $10^5$ rad/s the current will lag by $45^\circ$ w.r.t. voltage

Answer 1

Answer: (A), (B), (D)

A, B, and D

Answer 2

Explanation:

1. Calculate the current (I) at resonance:

   At resonance, the circuit behaves purely resistively. The voltage across the resistance ($V_R$) is given as 60 V and the resistance (R) is 120 $\Omega$. Using Ohm's law: $I = \frac{V_R}{R} = \frac{60 \, \text{V}}{120 \, \Omega} = 0.5 \, \text{A}$

2. Calculate the Inductance (L):

   At resonance, the voltage across the inductance ($V_L$) is given as 40 V. The inductive reactance ($X_L$) at resonance is $X_L = \omega_0 L$, where $\omega_0$ is the angular resonance frequency (4 × $10^5$ rad/s). $V_L = I \cdot X_L = I \cdot (\omega_0 L)$ $40 \, \text{V} = 0.5 \, \text{A} \cdot (4 \times 10^5 \, \text{rad/s} \cdot L)$ $L = \frac{40}{0.5 \times 4 \times 10^5} = \frac{40}{2 \times 10^5} = 20 \times 10^{-5} \, \text{H} = 2 \times 10^{-4} \, \text{H}$ Therefore, option A is correct.

3. Calculate the Capacitance (C):

   At resonance, the inductive reactance equals the capacitive reactance ($X_L = X_C$). $\omega_0 L = \frac{1}{\omega_0 C}$ $C = \frac{1}{\omega_0^2 L}$ Substitute the values of $\omega_0$ and L: $C = \frac{1}{(4 \times 10^5 \, \text{rad/s})^2 \times (2 \times 10^{-4} \, \text{H})}$ $C = \frac{1}{(16 \times 10^{10}) \times (2 \times 10^{-4})} = \frac{1}{32 \times 10^6} \, \text{F} = \frac{1}{32} \times 10^{-6} \, \text{F}$ Therefore, option B is correct.

4. Calculate the phase angle at angular frequency $\omega = 8 \times 10^5$ rad/s:

   First, calculate the new inductive reactance ($X_L'$) and capacitive reactance ($X_C'$): $X_L' = \omega L = (8 \times 10^5 \, \text{rad/s}) \times (2 \times 10^{-4} \, \text{H}) = 160 \, \Omega$ $X_C' = \frac{1}{\omega C} = \frac{1}{(8 \times 10^5 \, \text{rad/s}) \times (\frac{1}{32} \times 10^{-6} \, \text{F})}$ $X_C' = \frac{1}{(8/32) \times 10^{-1}} = \frac{1}{(1/4) \times 0.1} = \frac{1}{0.25 \times 0.1} = \frac{1}{0.025} = 40 \, \Omega$

   Now, calculate the phase angle ($\phi$) using the formula: $\tan \phi = \frac{X_L' - X_C'}{R}$ $\tan \phi = \frac{160 \, \Omega - 40 \, \Omega}{120 \, \Omega} = \frac{120 \, \Omega}{120 \, \Omega} = 1$ $\phi = \arctan(1) = 45^\circ$ Since $X_L' > X_C'$, the circuit is inductive, meaning the current lags the voltage. Therefore, at angular frequency 8 × $10^5$ rad/s, the current will lag by 45° w.r.t. voltage. Option D is correct, and option C is incorrect.