Question

Two capacitors A and B with capacities 3$\mu$F , 2$\mu$F respectively are charged to a potential difference of 100 V and 180 V respectively. The plates of the capacitors are connected as shown in the figure. With one wire from each capacitor free. The upper plate of A is positive and that of B is negative. An uncharged capacitor C with lead wires falls on the free ends to complete the circuit. Final charge on capacitor having capacitance $3\mu F$ will be :

• A. $90\mu C$
• B. $150\mu C$
• C. $210\mu C$
• D. $300\mu C$

Answer 1

The correct option is A $90\mu C$

$\Delta V_A+\Delta V_B=V_C$
$(100-\frac{Q}{3})+(180-\frac{Q}{2})=\frac{Q}{2}$
$⟹Q=210\mu C$
Final Charge on A = Initial Charge $-210\mu$ C $=300-210=90\mu C$