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: ![JEE Main 2015 Physics, Alternating Current — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Alternating_Current/648b5a64417cc3fb48d65ee6/question_1__q_648b5a64417cc3fb48d65ee6__cdn-question-pool.getmarks.app__3b085824-196f-41fc-b86d-ebc01080c60d-image__3df7fe9a57_final_ppt_sync.png) 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)

Comparing to the damped pendulum. We write, $\text{m}\frac{\text{d}\text{v}}{\text{d}\text{t}}=-\text{kx}-\text{bv}$, $bv$ is resistive force,$\Rightarrow$ Amplitude, $\text{A}={\text{A}}_{0}{\text{e}}^{-\frac{\text{bt}}{2\text{m}}}$. Comparing results, we can write $\frac{\text{q}}{\text{c}}=+\text{L}\frac{\text{d}\text{i}}{\text{d}\text{t}}+\text{iR}$, as charge decreasing, ![JEE Main 2015 Physics, Alternating Current — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Alternating_Current/648b5a64417cc3fb48d65ee6/solution_1__q_648b5a64417cc3fb48d65ee6__cdn-question-pool.getmarks.app__3b085824-196f-41fc-b86d-ebc01080c60d-image__3df7fe9a57_final_ppt_sync.png) $\frac{\text{q}}{\text{c}}=\text{L}\frac{{\text{d}}^{2}\text{q}}{\text{d}{\text{t}}^{2}}-\frac{\text{d}\text{q}}{\text{d}\text{t}}\text{R}$ $\Rightarrow \text{A}=\text{q}\text{; }\text{R}=\text{b}\text{, }\text{m}=\text{L}$ $\text{⇒}\text{q}={\text{q}}_{0}{\text{e}}^{-\frac{\text{Rt}}{2\text{L}}}$$\text{⇒}{\text{q}}^{2}={\text{q}}_{0}^{2}{\text{e}}^{-\frac{\text{Rt}}{\text{L}}}$$\text{⇒}{\text{q}}^{2}={\text{q}}_{0}^{2}{\text{e}}^{-\frac{\text{Rt}}{\text{L}}}$. Exponentially decreasing function, ${L}_{1}$ graph has higher values than ${L}_{2}$.

JEE Main 2015 — Physics Electromagnetism

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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:

JEE Main 2015 Physics, Alternating Current — question figure

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)

Official previous-year question

Held on 4 Apr 2015 · Verified 6 Jul 2026.

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Comparing to the damped pendulum. We write, $\text{m}\frac{\text{d}\text{v}}{\text{d}\text{t}}=-\text{kx}-\text{bv}$ , $bv$ is resistive force, $\Rightarrow$ Amplitude, $\text{A}={\text{A}}_{0}{\text{e}}^{-\frac{\text{bt}}{2\text{m}}}$ . Comparing results, we can write

$\frac{\text{q}}{\text{c}}=+\text{L}\frac{\text{d}\text{i}}{\text{d}\text{t}}+\text{iR}$ , as charge decreasing,

JEE Main 2015 Physics, Alternating Current — solution figure

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

$\Rightarrow \text{A}=\text{q}\text{; }\text{R}=\text{b}\text{, }\text{m}=\text{L}$

$\text{⇒}\text{q}={\text{q}}_{0}{\text{e}}^{-\frac{\text{Rt}}{2\text{L}}}$ $\text{⇒}{\text{q}}^{2}={\text{q}}_{0}^{2}{\text{e}}^{-\frac{\text{Rt}}{\text{L}}}$ $\text{⇒}{\text{q}}^{2}={\text{q}}_{0}^{2}{\text{e}}^{-\frac{\text{Rt}}{\text{L}}}$ .

Exponentially decreasing function, ${L}_{1}$ graph has higher values than ${L}_{2}$ .

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Solution

$\frac{\text{q}}{\text{c}}=+\text{L}\frac{\text{d}\text{i}}{\text{d}\text{t}}+\text{iR}$ , as charge decreasing,

JEE Main 2015 Physics, Alternating Current — solution figure

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

$\Rightarrow \text{A}=\text{q}\text{; }\text{R}=\text{b}\text{, }\text{m}=\text{L}$

Exponentially decreasing function, ${L}_{1}$ graph has higher values than ${L}_{2}$ .

JEE Main 2015 — Physics Electromagnetism

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JEE Main 2015 — Physics Electromagnetism

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JEE Main Electromagnetism in other years

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Mechanics

Modern Physics

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Waves & Oscillations

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