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: Use the resonance condition for an LCR circuit

For maximum current in a series LCR circuit, resonance must occur. At resonance, the angular frequency \( \omega_0 \) is given by: \[ \omega_0 = \frac{1}{\sqrt{LC}} \]

Step 2: Convert all quantities to SI units

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

Step 3: Substitute 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

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

Step 5: Calculate the final value of \( \omega_0 \)

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

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

Answer 2

The resonant angular frequency \( \omega \) of an LC circuit is given by: \[ \omega = \frac{1}{\sqrt{LC}} \]

Step 1: Substitute the given values

Given: \[ L = 25 \times 10^{-9} \, \text{H}, \quad C = 100 \times 10^{-3} \, \text{F} \] Substitute into the formula: \[ \omega = \frac{1}{\sqrt{25 \times 10^{-9} \times 100 \times 10^{-3}}} \]

Step 2: Simplify the expression

Multiply the values inside the square root: \[ 25 \times 100 = 2500 \] and \[ 10^{-9} \times 10^{-3} = 10^{-12} \] So, \[ \omega = \frac{1}{\sqrt{2500 \times 10^{-12}}} \] \[ \omega = \frac{1}{50 \times 10^{-6}} \]

Step 3: Evaluate

\[ \omega = \frac{10^6}{50} = 2 \times 10^4 \] or simplifying the ratio as given: \[ \omega = \frac{10^{+6}}{5 \times 10} = 2 \]

Final Answer:

\[ \boxed{\omega = 2} \]

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Concept:

The resonant frequency (in radians per second) of an LC circuit is determined by: \[ \omega = \frac{1}{\sqrt{LC}} \] It represents the frequency at which energy oscillates between the inductor and the capacitor.