
q1′=6×5=30μC Finally
6 VC+12 Vc=30+018 VC=30 VC=1830=35Volt⇒q1=36×5=10μC⇒q2=312×5=20μC
JEE Main 2025 — Physics Electromagnetism
A capacitor, C1=6μ F is charged to a potential difference of V0=5 V using a 5 V battery. The battery is removed and another capacitor, C2=12μ 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 (q1 and q2) on the capacitors C1 and C2 when equilibrium condition is reached. 
Held on 29 Jan 2025 · Verified 6 Jul 2026.
q1=10μC,q2=20μC
q1=30μC,q2=15μC
q1=20μC,q2=10μC
q1=15μC,q2=30μC
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