JEE Main 2015 · 4 Apr · Shift 1 · Physics Q30

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PhysicsSingle correct+4 / -1Question 30 of 90

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 ( $Q^2_{\text{Max}}$ ) 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)

A series loop of R, L and the capacitor C.

Both decay; the L₁ curve decays more slowly (lies above L₂)

Graph of Q²_max against t for L₁ and L₂: two decaying curves, L₁ above L₂.

Both decay; the L₂ curve decays more slowly (lies above L₁)

Graph of Q²_max against t for L₁ and L₂: two decaying curves, L₂ above L₁.

Both fall as straight lines, L₁ less steeply

Graph of Q²_max against t for L₁ and L₂: two straight falling lines, L₁ less steep than L₂.

A constant Q₀ for both L₁ and L₂

Graph of Q²_max against t for L₁ and L₂: a horizontal dashed line at Q₀ for both L₁ and L₂.

Answer and worked solution

Official answer: A

The charge amplitude decays as $Q_{max}=Q_0e^{-Rt/2L}$ , so $Q_{max}^2=Q_0^2e^{-Rt/L}$ . A larger L decays more slowly, so the $L_1$ curve lies above $L_2$ : graph (1).

Topics:LCR circuits and resonance Damped and forced oscillations

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JEE Main 2015 · 4 Apr · Shift 1, Physics question 30