An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to $\mathrm{Q}_{0}$ and then connected to the L and R as shown. If a student plots graphs of the square of maximum charge ( $\mathrm{Q}_{\text {max }}^{2}$ ) on the capacitor with time ( t ) for two different values $\mathrm{L}_{1}$ and $\mathrm{L}_{2}\left(\mathrm{~L}_{1}>\mathrm{L}_{2}\right)$ of L then which of the following represents this graph correctly? (Plots are schematic and not drawn to scale)

JEE Main · 2015 · 04 Apr 2015 · Offline · Electromagnetic Induction

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.
If a student plots graphs of the square of maximum charge (

Q_{max}^{2}

) 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)

Answer: (D)

When the charged capacitor is connected across

L

and

R

, the charge oscillates inside a decaying envelope

Q_{m a x} = Q_{0} e^{- R t /2 L}

, so

Q_{m a x}^{2} = Q_{0}^{2} e^{- R t / L}

falls exponentially from

Q_{0}^{2}

. The decay rate

R / L

is smaller for the larger inductance, so with

L_{1} > L_{2}

the

L_{1}

curve loses its energy more slowly and stays above the

L_{2}

curve at every instant, both starting together at

Q_{0}^{2}

. The correct plot is therefore the pair of exponentially decaying curves with

L_{1}

lying above

L_{2}

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option