Question

In a series LCR circuit, a resistor of \( 300 \, \Omega \), a capacitor of \( 25 \, \text{nF} \), and an inductor of \( 100 \, \text{mH} \) are used. For maximum current in the circuit, the angular frequency of the AC source is -----\( \times 10^4 \) radians s\(^{-1}\).

Hint:

The angular frequency for resonance in an LCR circuit is given by \( \omega_0 = \frac{1}{\sqrt{LC}} \), where \( L \) is the inductance and \( C \) is the capacitance.

Answer 1

Correct Answer: 2

Step 1: Apply the resonance condition for an LCR circuit

Maximum current in a series LCR circuit is achieved at resonance. The angular frequency \( \omega_0 \) at resonance is defined as: \[ \omega_0 = \frac{1}{\sqrt{LC}} \]

Step 2: Convert all component values to SI units

• Inductance: \( L = 100\,\text{mH} = 100 \times 10^{-3}\,\text{H} = 0.1\,\text{H} \)
• Capacitance: \( C = 25\,\text{nF} = 25 \times 10^{-9}\,\text{F} = 2.5 \times 10^{-8}\,\text{F} \)

Step 3: Substitute the SI unit values into the resonance formula

\[ \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1 \times 2.5 \times 10^{-8}}} \] \[ = \frac{1}{\sqrt{2.5 \times 10^{-9}}} \]

Step 4: Simplify the square root term

\[ \sqrt{2.5 \times 10^{-9}} = \sqrt{2.5} \times \sqrt{10^{-9}} = 1.58 \times 10^{-4.5} \]

Step 5: Compute the final angular frequency \( \omega_0 \)

\[ \omega_0 = \frac{1}{1.58 \times 10^{-4.5}} = \frac{1}{1.58 \times 3.16 \times 10^{-5}} \approx \frac{1}{5 \times 10^{-5}} = 2 \times 10^4\,\text{rad/s} \]

Final Answer: The angular frequency at resonance is \( \boxed{2 \times 10^4\,\text{rad/s}} \).