JEE MAIN - Physics (2015 (Offline) - No. 5)
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 :
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If a student plots graphs of the square of maximum charge $$\left( {Q_{Max}^2} \right)$$ on the capacitor with time$$(t)$$ for two different values $${L_1}$$ and $${L_2}$$ $$\left( {{L_1} > {L_2}} \right)$$ of $$L$$ then which of the following represents this graph correctly ?
$$\left( {plots\,\,are\,\,schematic\,\,and\,\,niot\,\,drawn\,\,to\,\,scale} \right)$$
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Explanation
From $$KVL$$ at any time $$t$$
_en_5_6.png)
$${q \over c} - iR - L{{di} \over {dt}} = 0$$
$$i = - {{dq} \over {dt}} \Rightarrow {q \over c} + {{dq} \over {dt}}R + {{L{d^2}q} \over {d{t^2}}} = 0$$
$${{{d^2}q} \over {d{t^2}}} + {R \over L}{{dq} \over {dt}} + {q \over {Lc}} = 0$$
From damped harmonic oscillator, the amplitude is
given by $$A = {A_0}e - {{dt} \over {2m}}$$
Double differential equation
$${{{d^2}x} \over {d{t^2}}} + {b \over m}{{dx} \over {dt}} + {k \over m}x = 0$$
$${Q_{\max }} = {Q_0}e - {{Rt} \over {2L}} \Rightarrow Q_{\max }^2 = Q_0^2e - {{Rt} \over L}$$
Hence damping will be faster for lesser self inductance.
$${q \over c} - iR - L{{di} \over {dt}} = 0$$
$$i = - {{dq} \over {dt}} \Rightarrow {q \over c} + {{dq} \over {dt}}R + {{L{d^2}q} \over {d{t^2}}} = 0$$
$${{{d^2}q} \over {d{t^2}}} + {R \over L}{{dq} \over {dt}} + {q \over {Lc}} = 0$$
From damped harmonic oscillator, the amplitude is
given by $$A = {A_0}e - {{dt} \over {2m}}$$
Double differential equation
$${{{d^2}x} \over {d{t^2}}} + {b \over m}{{dx} \over {dt}} + {k \over m}x = 0$$
$${Q_{\max }} = {Q_0}e - {{Rt} \over {2L}} \Rightarrow Q_{\max }^2 = Q_0^2e - {{Rt} \over L}$$
Hence damping will be faster for lesser self inductance.
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