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 one 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. Charge flow through section 2 in the direction as shown in figure is :

• A. $210\mu C$
• B. $150\mu C$
• C. $-210\mu C$
• D. $-150\mu C$

Answer 1

The correct option is C $-210\mu C$

Let Q be the charge flowing through the circuit . When the switch is pressed , the voltages developed on capacitors A,B and C are
$V_A=\frac{Q}{3\times {10}^{-6}}$ volt , $V_B=\frac{Q}{2\times {10}^{-6}}$ volt and $V_C=\frac{Q}{2\times {10}^{-6}}$ volt
Applying Kirchoff's law to the loop , we have
$\Delta V_A+\Delta V_B+\Delta V_C=0$
$-(100-\frac{Q}{3\times {10}^{-6}})-(180-\frac{Q}{2\times {10}^{-6}})+\frac{Q}{2\times {10}^{-6}}=0$
which gives $Q=210\times {10}^{-6}=210\mu C$