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A capacitor is charged to an initial charge Q0 and then connected across an inductor L and resistor R in series, forming a damped LCR oscillator. A student plots the square of the maximum (envelope) charge on the capacitor, Q_max², against time t for two different inductance values L1 and L2 with L1 > L2 (R and C kept the same). Which statement correctly describes the two curves?

1. Both curves decay exponentially from Q0²; the curve for L2 (smaller L) decays faster than the curve for L1.
2. Both curves decay exponentially from Q0²; the curve for L1 (larger L) decays faster than the curve for L2.
3. Both curves are straight lines decreasing with time, independent of L.
4. Q_max² increases with time for both inductances.

Correct answer: Both curves decay exponentially from Q0²; the curve for L2 (smaller L) decays faster than the curve for L1.

Solution

In an underdamped series LCR circuit the charge oscillates with an exponentially decaying envelope Q_max(t) = Q0*exp(-R*t/(2L)). Squaring gives Q_max² = Q0²*exp(-R*t/L), an exponential decay starting from Q0² with decay constant R/L. A larger L gives a smaller decay constant (slower decay), while a smaller L gives a larger decay constant (faster decay). Since L1 > L2, the L1 curve decays more slowly and the L2 curve decays faster; both start at Q0².

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