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Exercises · 7.6

Q.A charged 30 μF30\ \mu\text{F} capacitor is connected to a 27 mH27\ \text{mH} inductor. What is the angular frequency of free oscillations of the circuit?

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The circuit is an ideal LC oscillator. The angular frequency of free oscillations depends only on LL and CC via ω=1/LC\omega = 1/\sqrt{LC}. Substituting the given values gives ω≈1.11×103 rad/s\omega \approx 1.11 \times 10^3\ \text{rad/s}.

The Concept: Why an LC Circuit Oscillates

When you connect a charged capacitor to an inductor, you create a perfect electrical pendulum. The capacitor stores energy in its electric field; the inductor stores energy in its magnetic field. There is no resistor here, so no energy is lost — the circuit will oscillate forever at a single natural frequency.

The key insight is that this oscillation is analogous to a mass on a spring. In a mechanical system, the angular frequency is ω=k/m\omega = \sqrt{k/m} where kk is the spring constant and mm is the mass. In an LC circuit, the inductor LL plays the role of inertia (mass), and the capacitor CC plays the role of stiffness (the reciprocal of the spring constant). So the natural angular frequency is:

ω=1LC\omega = \frac{1}{\sqrt{LC}}

This is one of the most fundamental results in AC circuit theory. It tells you that the oscillation frequency depends only on the component values, not on how much charge you started with or what the initial voltage was.

Step-by-Step Solution

1. Identify the circuit type.

We have only a capacitor and an inductor — no resistor. This is an ideal LC circuit (also called a tank circuit). Free oscillations means the circuit is left to itself after the initial energy is supplied (here, by charging the capacitor).

2. Recall the formula for angular frequency.

For an LC circuit, the charge on the capacitor and the current in the inductor both vary sinusoidally with time. The angular frequency ω\omega (in radians per second) is:

ω=1LC\omega = \frac{1}{\sqrt{LC}}

Watch out

A common mistake is to confuse angular frequency ω\omega with ordinary frequency ff. They are related by ω=2πf\omega = 2\pi f, but the question explicitly asks for angular frequency, so we use the formula above directly — no extra factor of 2π2\pi needed.

3. Write down the given values with correct units.

  • Capacitance: C=30 μF=30×10−6 FC = 30\ \mu\text{F} = 30 \times 10^{-6}\ \text{F}
  • Inductance: L=27 mH=27×10−3 HL = 27\ \text{mH} = 27 \times 10^{-3}\ \text{H}

Always convert micro and milli to the base SI units before plugging in.

4. Substitute into the formula.

ω=1(27×10−3)(30×10−6)\omega = \frac{1}{\sqrt{(27 \times 10^{-3})(30 \times 10^{-6})}}

First, multiply the numbers inside the square root:

L×C=27×30×10−3×10−6=810×10−9=8.10×10−7L \times C = 27 \times 30 \times 10^{-3} \times 10^{-6} = 810 \times 10^{-9} = 8.10 \times 10^{-7}

So: …

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