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  • A L – C – R circuit is equivalent to a damped pendulum. In an L – C – R circuit the capacitor is charged to <img alt="\mathrm{Q}_0" src="https://entrancecorner.oncodecogs.co

A L – C – R circuit is equivalent to a damped pendulum. In an L – C – R circuit the capacitor is charged to \mathrm{Q}_0 and then connected to the L and R as shown below.
                                               
If a student plots graphs of the square of maximum charge \mathrm{\left(Q_{\max }^2\right)} on the capacitor with time (t) for two different values L1 and L ( \mathrm{\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):

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)

best_answer

\text { At any time } t \text { apply } K V L, \frac{q}{C}-i R-L \frac{d i}{d t}=0
   

\begin{array}{r} \mathrm{i}=-\frac{\mathrm{dq}}{\mathrm{dt}} \\ \Rightarrow \frac{\mathrm{q}}{\mathrm{C}}+\frac{\mathrm{dq}}{\mathrm{dt}} \mathrm{R}+\frac{\mathrm{Ld}^2 \mathrm{q}}{\mathrm{dt}^2}=0 \\ \Rightarrow \frac{\mathrm{d}^2 \mathrm{q}}{\mathrm{dt}^2}+\frac{\mathrm{R}}{\mathrm{L}} \frac{\mathrm{dq}}{\mathrm{dt}}+\frac{\mathrm{q}}{\mathrm{LC}}=0 \end{array}

from damped harmonic oscillator, the amplitude is given by 
\mathrm{A}=\mathrm{A}_0 \mathrm{e}^{-\frac{\mathrm{k}}{2 \mathrm{~m}}},
for general equation of double differential equation 
\frac{\mathrm{d}^2 \mathrm{x}}{\mathrm{dt}^2}+\frac{\mathrm{b}}{\mathrm{m}} \frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\mathrm{k}}{\mathrm{m}} \mathrm{x}=0
\begin{aligned} & \Rightarrow \mathrm{Q}_{\max }^{(\mathrm{t})}=\mathrm{Q}_0 \mathrm{e}^{-\frac{\mathrm{Rt}}{2 \mathrm{~L}}} \\ & \Rightarrow \mathrm{Q}_{\max }^{(\mathrm{t})}=\mathrm{Q}_0^2 \mathrm{e}^{-\frac{\mathrm{Rt}}{2 \mathrm{~L}}} \end{aligned}
Lesser the self-inductance, faster will be damping hence.

Posted by

Sanket Gandhi

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