Question

A series L-C-R circuit containing a resistance of $120 \; \Omega$ has angular frequency $4 \times 10^5 \text{ rad s}^{-1}$. At resonance, the voltage across the resistance and inductor are $60 \text{ V}$ and $40 \text{ V}$ respectively, then the value of inductance will be

• A. $0.2 \text{ mH}$
• B. $0.4 \text{ mH}$
• C. $0.8 \text{ mH}$
• D. $0.6 \text{ mH}$

Hint:

At resonance, you can utilize the quality factor relationship shortcut directly: $\frac{V_L}{V_R} = \frac{\omega L}{R}$. Substituting the values directly gives $\frac{40}{60} = \frac{4 \times 10^5 \cdot L}{120}$, which easily simplifies to isolated $L = 0.2 \text{ mH}$ in one single algebraic step.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem presents an alternating current series L-C-R circuit operating at its resonant frequency. Given the resistance value, angular frequency, voltage across the resistor, and voltage across the inductor, we need to calculate the value of the inductance $L$.

Step 2: Key Formula or Approach:
At resonance in a series L-C-R circuit, the current $I$ is limited solely by the resistance $R$ because the inductive and capacitive reactances cancel out. By Ohm's law, the circuit current is:
$$I = \frac{V_R}{R}$$ The potential drop across the inductor $V_L$ depends on its inductive reactance $X_L$:
$$V_L = I \cdot X_L = I \cdot (\omega L)$$

Step 3: Detailed Explanation:
Let's list the given variables:
$R = 120 \; \Omega$
$\omega = 4 \times 10^5 \text{ rad s}^{-1}$
$V_R = 60 \text{ V}$
$V_L = 40 \text{ V}$
First, calculate the current $I$ flowing through the circuit loop using the resistor's parameters:
$$I = \frac{60 \text{ V}}{120 \; \Omega} = 0.5 \text{ A}$$ Since it is a series configuration, the same current of $0.5 \text{ A}$ passes through the inductor. Use this current in the inductor voltage equation:
$$40 = 0.5 \times (4 \times 10^5 \times L)$$ $$40 = (2 \times 10^5) \cdot L$$ Isolating $L$ by dividing both sides by $2 \times 10^5$:
$$L = \frac{40}{2 \times 10^5} = 20 \times 10^{-5} = 2 \times 10^{-4} \text{ H}$$ Converting Henrys to millihenrys ($1 \text{ mH} = 10^{-3} \text{ H}$):
$$L = 0.2 \times 10^{-3} \text{ H} = 0.2 \text{ mH}$$

Step 4: Final Answer:
The inductance value is $0.2 \text{ mH}$, which corresponds to option (A).