Question

A series LCR circuit containing a 5.0 H inductor, 80 µF capacitor and 40 Ω resistor is connected to a 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be

• A. 42 rad/s and 58 rad/s
• B. 25 rad/s and 75 rad/s
• C. 50 rad/s and 25 rad/s
• D. 46 rad/s and 54 rad/s

Answer 1

The Correct Option is D

In this problem, we have an RLC series circuit with a given inductor, capacitor, and resistor, connected to an AC source. Let's solve for the angular frequencies where the power transferred to the circuit is half of the power at resonance.

We are given:

• Inductance, \(L = 5.0 \,\text{H}\)
• Capacitance, \(C = 80 \, \mu\text{F} = 80 \times 10^{-6} \,\text{F}\)
• Resistance, \(R = 40 \, \Omega\)
• AC Source Voltage, \(V = 230 \,\text{V}\)

First, calculate the resonant angular frequency \(\omega_0\) using the formula:

\(\omega_0 = \frac{1}{\sqrt{LC}}\)

Substitute the given values:

\(\omega_0 = \frac{1}{\sqrt{5.0 \times 80 \times 10^{-6}}}\)

\(\omega_0 = \frac{1}{\sqrt{4 \times 10^{-4}}} = \frac{1}{0.02} = 50 \,\text{rad/s}\)

The power at resonance is given by:

\(P = \frac{V^2}{R}\)

At angular frequencies where power is half of the resonant power, the relation is:

\(\omega = \omega_0 \pm \frac{R}{2L}\)

Calculate \(\frac{R}{2L}\):

\(\frac{R}{2L} = \frac{40}{2 \times 5} = 4 \,\text{rad/s}\)

Thus, the required angular frequencies are:

• \(\omega_1 = \omega_0 - \frac{R}{2L} = 50 - 4 = 46 \,\text{rad/s}\)
• \(\omega_2 = \omega_0 + \frac{R}{2L} = 50 + 4 = 54 \,\text{rad/s}\)

Therefore, the angular frequencies at which the power transferred to the circuit is half the power at the resonant angular frequency are 46 rad/s and 54 rad/s.

Hence, the correct answer is:

46 rad/s and 54 rad/s