Question

Two capacitors $C_{1}$ and $C_{2}$ in a circuit are joined as shown in figure. The potentials of points A and B are $V_{1}$ and $V_{2}$ respectively. Then the potential of point D will be

• A. $\frac{\left(V_{1}+V_{2}\right)}{2}$
• B. $\frac{C_{2} V_{1} +C _{1}V_{2}}{C_{1}+C_{2}}$
• C. $\frac{C_{1}V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}$
• D. $\frac{C_{2}V_{1}+C_{1} V_{2}}{C_{1}+C_{2}}$

Answer 1

Answer: (C)

$\frac{C_{1}V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}$

Answer 2

Consider the potential at D be ?V?.
Potential drop across $C_{1} is \, \left(V-V_{1}\right)$ and $C_{2} \, is\, \left(V_{2}-V\right)$
$\therefore\quad q_{1} =C_{1} \left(V-V_{1}\right), q_{2} =C_{2} \left(V_{2}-V\right)$
$As\, q_{1} =q_{2} \quad$ [capacitors are in series]
$\therefore\quad C_{1} \left(V-V_{1}\right)=C_{2} \left(V_{2}-V\right)$
$V=\frac{C_{1}V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}$