Question

In the given figure, capacitors $C_1=1\mu F$ and $C_2=2\mu F$ are charged to a potential difference of $100V$ but with opposite polarity as shown. Switches $S_{1}and{S}_2$ are now closed. What is the potential difference between points $a$ and $b$?

• A. $200V$
• B. $100V$
• C. $33V$
• D. $25V$

Answer 1

The correct option is C $33V$

When switches are closed, $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{1\times 2}{1+2}=2/3\mu C$ and $Q_{eq}=C_{eq}V=2/3\times 100=200/3\mu C$

As they are in series so charge on each capacitors is equal to $Q_{eq}$

$V_1=\frac{Q_{eq}}{C_1}=200/3$ and $V_2=\frac{Q_{eq}}{C_2}=100/3$

As they are in opposite polarity, the potential difference between a and b will be $V^{\prime }=V_1-V_2=200/3-100/3\sim 33V$

When switches are closed, $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\frac{1\times 2}{1+2}=2/3\mu C$ and $Q_{eq}=C_{eq}V=2/3\times 100=200/3\mu C$

As they are in series so charge on each capacitors is equal to $Q_{eq}$

$V_1=\frac{Q_{eq}}{C_1}=200/3$ and $V_2=\frac{Q_{eq}}{C_2}=100/3$

As they are in opposite polarity, the potential difference between a and b will be $V^{\prime }=V_1-V_2=200/3-100/3\sim 33V$