Two condensers C_1 and C_2 in a circuit are&n | vedclass

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Question

$V_{1}-V_{2}$ will be divided between $C_{1}$ and $C_{2}$ and $V_{D}$ is potential at point $D$ in series.

Potential difference across $C_{1}$ is

Vedclass · Accepted Answer

$\frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}$

Vedclass · Answer

$V_{1}-V_{2}$ will be divided between $C_{1}$ and $C_{2}$ and $V_{D}$ is potential at point $D$ in series.

Potential difference across $C_{1}$ is

$V_{1}-V_{D}=\frac{C_{2}\left(V_{1}-V_{2}\right)}{C_{1}+C_{2}}$

$V_{D}=V_{1}-\frac{C_{2}\left(V_{1}-V_{2}\right)}{C_{1}+C_{2}}$

or $V_{D}=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}$

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 $D$ will be

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