Three circuits, each consisting of a switch ‘S’ and two capacitors, are initially charged, as shown in the figure. After the switch has been closed, in which circuit will the charge on the left-hand capacitor (i) increase, (ii) decrease and (iii) remain same? Give reasons.

Charge flows in the circuit until both the capacitors are at same potential. We know, that the potential, V = q/c
Vleft = 6Q/2C = 3Q/C and Vright =3Q/C
⇒ Vleft = Vright
Therefore, voltage in left capacitor stays the same.
Vleft = 6Q/C and Vright = 3Q/C
⇒ Vleft > Vright
Therefore, voltage in left capacitor will decrease.
Vleft = 6Q/3C = 2Q/C and Vright = 3Q/C
⇒ Vleft < Vright
Therefore, voltage in left capacitor will increase
Couldn't generate an explanation.
Generated by AI. May contain inaccuracies — always verify with your textbook.