Physics Circle home Student Login

Home›Chapters›Capacitors PYQs›Charge Sharing & Loss of Energy›Question 9

JEE Main 2025 (29 Jan, Shift 2) · Charge Sharing & Loss of Energy

Charge Sharing & Loss of Energy question from JEE Main 2025 (29 Jan, Shift 2)

Q9JEE Main 2025 · 29 JAN · SHIFT 2Medium

Charge Sharing & Loss of Energy

A capacitor, C1=6 μFC_1 = 6\ \mu\text{F} is charged to a potential difference of V0=5 VV_0 = 5\text{ V} using a 5V battery. The battery is removed and another capacitor, C2=12 μFC_2 = 12\ \mu\text{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 (q1q_1 and q2q_2) on the capacitors C1C_1 and C2C_2 when equilibrium condition is reached.

Figure

Show answer and solution

Answer: (a)

  1. Q=C1V0=6×5=30 μQ = C_1V_0 = 6 \times 5 = 30\ \muC
  2. Common potential V=306+12=53V = \frac{30}{6 + 12} = \frac{5}{3} V
  3. q1=6×53=10 μq_1 = 6 \times \frac{5}{3} = 10\ \muC, q2=12×53=20 μq_2 = 12 \times \frac{5}{3} = 20\ \muC
  4. So the answer is (a).
Report a mistake

More questions on Charge Sharing & Loss of Energy