An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q₀ and then connected to the L and R as shown below: If a student plots graphs of the square of maximum charge (QMax²) on the capacitor with time (t) for two different values L₁ and L₂ (L₁ > L₂) of L then which of the following represents this graph correctly? (plots are schematic and not drawn to scale) (a) (b) (c) (d)

Answer: (c). tQ²maxL1 (larger)L2 For a damped LCR circuit, the charge amplitude is Qmax = Q₀ e-Rt/2L. So Qmax² = Q₀² e-Rt/L: an exponential decay, not a straight line or a constant. A larger L gives slower decay, so the L₁ curve ( L₁ > L₂) stays above the L₂ curve. So the answer is (c).

An LCR circuit is equivalent to a damped pendulum. In an LCR circuit… | JEE Main 2015

Q20JEE Main 2015Difficult

LC Oscillations

An $LCR$ circuit is equivalent to a damped pendulum. In an $LCR$ circuit the capacitor is charged to $Q_0$ and then connected to the $L$ and $R$ as shown below:

If a student plots graphs of the square of maximum charge $\left(Q_{Max}^2\right)$ on the capacitor with time $(t)$ for two different values $L_1$ and $L_2\ (L_1 > L_2)$ of $L$ then which of the following represents this graph correctly?

(plots are schematic and not drawn to scale)

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Answer: (c)

For a damped LCR circuit, the charge amplitude is $Q_{max} = Q_0 e^{-Rt/2L}$ .
So $Q_{max}^2 = Q_0^2 e^{-Rt/L}$ : an exponential decay, not a straight line or a constant.
A larger $L$ gives slower decay, so the $L_1$ curve ( $L_1 > L_2$ ) stays above the $L_2$ curve.
So the answer is (c).

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