Question

Two capacitors $C_1=\mu F$ and $C_2=3\mu F$ each is charged to a potential difference of $100V$ but with opposite polarity as shown in the figure. When the switch S is closed, the new potential difference between the points a and b is :

• A. $200V$
• B. $100V$
• C. $50V$
• D. $25V$

Answer 1

The correct option is C $50V$

When S is closed, the capacitor are connected in parallel with oppsite polarity.
So total charge $Q_t=C_{2}V-C_{1}V=(3-1)\left(100\right)=200\mu C$
and total capacitance is $C_t=C_1+C_2=1+3=4\mu F$
Therefore common potential is $V_c=\frac{Q_t}{C_t}=\frac{200}{4}=50V$