In the given circuits $(a)$ and $(b)$, switches ${\text{S}}_{1}$ and ${\text{S}}_{2}$ are closed at $\text{t}=0$ and kept close for a long time. The variation of currents in the two circuits for $\text{t}\geq 0$ are shown in the options. (Figures are schematic and not drawn to scale.) ![JEE Main 2015 Physics, Capacitance — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Capacitance/648b5a64417cc3fb48d65f43/question_1__q_648b5a64417cc3fb48d65f43__cdn-question-pool.getmarks.app__7ffa1d61-53d0-417e-9169-ec7910377604-D716__1bcec33a95_final_ppt_sync.png)

In C-R series circuit, $\text{q}={\text{q}}_{0} (1-{\text{e}}^{\frac{-t}{\text{τ}}})$ $\Longrightarrow \text{q}=\text{CE} (1-{\text{e}}^{\frac{-\text{t}}{\text{RC}}})$ $\therefore \text{I}=\frac{\text{d}\text{q}}{\text{d}\text{t}}$ $=\frac{\text{CE}}{\text{RC}}(+{\text{e}}^{\frac{-\text{t}}{\text{RC}}})$ $\text{I}=\frac{\text{E}}{\text{R}} {\text{e}}^{\frac{-\text{t}}{\text{RC}}}$ $\Longrightarrow$ the urrent decays exponentially. In $L-R$ series circuit, $\text{I}={\text{I}}_{0}(1-{\text{e}}^{\frac{-\text{t}}{\text{τ}}})$ $\Rightarrow {\text{I}}_{0}=\frac{\text{E}}{\text{R}}$ $\text{⇒}\text{I}=\frac{\text{E}}{\text{R}} (1-{\text{e}}^{\frac{-\text{Rt}}{\text{L}}})$ $\Longrightarrow$ current grows exponentially. $\therefore$ For $C-R$ circuit, ![JEE Main 2015 Physics, Capacitance — solution figure 1](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Capacitance/648b5a64417cc3fb48d65f43/solution_1__q_648b5a64417cc3fb48d65f43__cdn-question-pool.getmarks.app__9244094e-cb65-45eb-8867-562090ade56d-image__8b10d4f663_final_ppt_sync.png) For $L-R$ circuit, ![JEE Main 2015 Physics, Capacitance — solution figure 2](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Capacitance/648b5a64417cc3fb48d65f43/solution_2__q_648b5a64417cc3fb48d65f43__cdn-question-pool.getmarks.app__image41-600daa47-8e7e-4222-a603-d91d5160ff6b__5b23e50084_final_ppt_sync.jpg)

JEE Main 2015 — Physics Electromagnetism

2015mcqhard

In the given circuits $(a)$ and $(b)$ , switches ${\text{S}}_{1}$ and ${\text{S}}_{2}$ are closed at $\text{t}=0$ and kept close for a long time. The variation of currents in the two circuits for $\text{t}\geq 0$ are shown in the options. (Figures are schematic and not drawn to scale.)

JEE Main 2015 Physics, Capacitance — question figure

Official previous-year question

Held on 10 Apr 2015 · Verified 6 Jul 2026.

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Solution

In C-R series circuit,

$\text{q}={\text{q}}_{0} (1-{\text{e}}^{\frac{-t}{\text{τ}}})$

$\Longrightarrow \text{q}=\text{CE} (1-{\text{e}}^{\frac{-\text{t}}{\text{RC}}})$

$\therefore \text{I}=\frac{\text{d}\text{q}}{\text{d}\text{t}}$

$=\frac{\text{CE}}{\text{RC}}(+{\text{e}}^{\frac{-\text{t}}{\text{RC}}})$

$\text{I}=\frac{\text{E}}{\text{R}} {\text{e}}^{\frac{-\text{t}}{\text{RC}}}$

$\Longrightarrow$ the urrent decays exponentially. In $L-R$ series circuit,

$\text{I}={\text{I}}_{0}(1-{\text{e}}^{\frac{-\text{t}}{\text{τ}}})$

$\Rightarrow {\text{I}}_{0}=\frac{\text{E}}{\text{R}}$

$\text{⇒}\text{I}=\frac{\text{E}}{\text{R}} (1-{\text{e}}^{\frac{-\text{Rt}}{\text{L}}})$

$\Longrightarrow$ current grows exponentially.

$\therefore$ For $C-R$ circuit,

JEE Main 2015 Physics, Capacitance — solution figure 1

For $L-R$ circuit,

JEE Main 2015 Physics, Capacitance — solution figure 2

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Solution

In C-R series circuit,

$\text{q}={\text{q}}_{0} (1-{\text{e}}^{\frac{-t}{\text{τ}}})$

$\Longrightarrow \text{q}=\text{CE} (1-{\text{e}}^{\frac{-\text{t}}{\text{RC}}})$

$\therefore \text{I}=\frac{\text{d}\text{q}}{\text{d}\text{t}}$

$=\frac{\text{CE}}{\text{RC}}(+{\text{e}}^{\frac{-\text{t}}{\text{RC}}})$

$\text{I}=\frac{\text{E}}{\text{R}} {\text{e}}^{\frac{-\text{t}}{\text{RC}}}$

$\Longrightarrow$ the urrent decays exponentially. In $L-R$ series circuit,

$\text{I}={\text{I}}_{0}(1-{\text{e}}^{\frac{-\text{t}}{\text{τ}}})$

$\Rightarrow {\text{I}}_{0}=\frac{\text{E}}{\text{R}}$

$\text{⇒}\text{I}=\frac{\text{E}}{\text{R}} (1-{\text{e}}^{\frac{-\text{Rt}}{\text{L}}})$

$\Longrightarrow$ current grows exponentially.

$\therefore$ For $C-R$ circuit,

JEE Main 2015 Physics, Capacitance — solution figure 1

For $L-R$ circuit,

JEE Main 2015 Physics, Capacitance — solution figure 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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Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

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