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 $(Q_{M a x}^{2})$ 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)

Question

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_{M a x}^{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)

Aicharya · Accepted Answer

Comparing to the damped pendulum. We write,

m \frac{d v}{d t} = - kx - bv

,

b v

is resistive force,

\Rightarrow

Amplitude,

A = A_{0} e^{- \frac{bt}{2 m}}

. Comparing results, we can write

\frac{q}{c} = + L \frac{d i}{d t} + iR

, as charge decreasing,

\frac{q}{c} = L \frac{d^{2} q}{d t^{2}} - \frac{d q}{d t} R

\Rightarrow A = q; R = b, m = L

⇒ q = q_{0} e^{- \frac{Rt}{2 L}}

⇒ q^{2} = q_{0}^{2} e^{- \frac{Rt}{L}}

⇒ q^{2} = q_{0}^{2} e^{- \frac{Rt}{L}}

.

Exponentially decreasing function,

L_{1}

graph has higher values than

L_{2}

.

An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q 0 and…