Question

Two condensers $C_{1}$ and $C_{2}$ in a circuit are joined as shown in figure. The potential of point $A$ is $V _{1}$ and that of $B$ is $V _{2}$. The potential of point $O$ will be

• A. $C _{1} V _{1}+ C _{2} V _{2}$
• B. $\frac{ C _{2} V _{1}+ C _{1} V _{1}}{ C _{1}+2 C _{2}}$
• C. $\frac{ C _{1} V _{1}+ C _{2} V _{2}}{ C _{1}+ C _{2}}$
• D. $\frac{V_{1}+V_{2}}{2}$

Answer 1

Answer: (C)

Both capacitors are joined in series combination in which charge remain same in all capacitors
So, charge on $C_{1}$ = charge on $C_{2}$
$\therefore C_{1} \Delta V =C_{2} \Delta V$
$\therefore C_{1}\left(V_{A}-V_{\Delta}\right) =C_{2}\left(V_{\Delta}-V_{B}\right) $
$C_{1}\left(V_{1}-V_{\Delta}\right)=C_{2}\left(C_{\Delta}-V_{2}\right)$
$\therefore C_{1} V_{1}-C_{1} V_{\Delta}=C_{2} V_{\Delta}-C_{2} V_{2}$
$\therefore C_{1} V_{1}+C_{2} V_{2}=C_{2} V_{\Delta}+C_{1} V_{\Delta}$
$V_{\Delta}=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}$
It means there is a common potential at point $O$