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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2025 · 29 January · Evening Shift · Physics · Capacitance

A capacitor, C_1=6 μ ~F is charged to a potential difference of...

A capacitor, C1=6μ FC_1=6 \mu \mathrm{~F} is charged to a potential difference of V0=5 V\mathrm{V}_0=5 \mathrm{~V} using a 5 V battery. The battery is removed and another capacitor, C2=12μ F\mathrm{C}_2=12 \mu \mathrm{~F} is inserted in place of the battery. When the switch 'S' is closed, the charge flows between the capacitors for some time until equilibrium condition is reached. What are the charges (q1andq2)onthecapacitors\left(q_1\right. and \left.q_2\right) on the capacitorsC_1 and C_2whenequilibriumconditionisreached.<imgsrc="https://cdn−question−pool.getmarks.app/pyq/jeemain/NH5mxSaPb5faWgINTjg−CiZmkugfPVfAxQ−dLBGf4Lo.original.fullsize.png"/><br/>when equilibrium condition is reached. <img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/NH5mxSaPb5faWgINTjg-CiZmkugfPVfAxQ-dLBGf4Lo.original.fullsize.png"/><br/>

Options

  1. Aq1=10μC,q2=20μCq_1=10 \mu \mathrm{C}, \mathrm{q}_2=20 \mu \mathrm{C}
  2. Bq1=30μC,q2=15μC\mathrm{q}_1=30 \mu \mathrm{C}, \mathrm{q}_2=15 \mu \mathrm{C}
  3. Cq1=20μC,q2=10μC\mathrm{q}_1=20 \mu \mathrm{C}, \mathrm{q}_2=10 \mu \mathrm{C}
  4. Dq1=15μC,q2=30μC\mathrm{q}_1=15 \mu \mathrm{C}, \mathrm{q}_2=30 \mu \mathrm{C}

Correct answer

A. q1=10μC,q2=20μCq_1=10 \mu \mathrm{C}, \mathrm{q}_2=20 \mu \mathrm{C}

Step-by-step solution

A capacitor, C_1=6 μ ~F is charged to a potential difference of V_0=5 ~V using a 5 V battery. The battery is removed
q1′=6×5=30μC\mathrm{q}_1^{\prime}=6 \times 5=30 \mu \mathrm{C} FinallyA capacitor, C_1=6 μ ~F is charged to a potential difference of V_0=5 ~V using a 5 V battery. The battery is removed
6 VC+12 Vc=30+018 VC=30 VC=3018=53Volt⇒q1=6×53=10μC⇒q2=12×53=20μC\begin{aligned} & 6 \mathrm{~V}_{\mathrm{C}}+12 \mathrm{~V}_{\mathrm{c}}=30+0 \\ & 18 \mathrm{~V}_{\mathrm{C}}=30 \\ & \mathrm{~V}_{\mathrm{C}}=\frac{30}{18}=\frac{5}{3} \mathrm{Volt} \\ & \Rightarrow \mathrm{q}_1=\frac{6 \times 5}{3}=10 \mu \mathrm{C} \\ & \Rightarrow \mathrm{q}_2=\frac{12 \times 5}{3}=20 \mu \mathrm{C}\end{aligned}
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