Q92 / 120 · VITEEE 2017 · Physics

Two capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ in a circuit are joined as shown in figure. The potentials of points $A$ and $\mathrm{B}$ are $\mathrm{V}_{1}$ and $\mathrm{V}_{2}$ respectively. Then the potential of point D will be Figure

  • A
    $\frac{\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right)}{2}$
  • B
    $\frac{\mathrm{C}_{2} \mathrm{~V}_{1}+\mathrm{C}_{1} \mathrm{~V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
  • Correct answer
    C
    $\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}$
  • D
    $\frac{\mathrm{C}_{2} \mathrm{~V}_{1}+\mathrm{C}_{1} \mathrm{~V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$

Solution

Consider the potential at D be 'V'. Potential drop across $C_{1}$ is $\left(V-V_{1}\right)$ and $C_{2}$ $\mathrm{i}$ $\left(\mathrm{V}_{2}-\mathrm{V}\right)$ $\therefore \mathrm{q}_{1}=\mathrm{C}_{1}\left(\mathrm{~V}-\mathrm{V}_{1}\right), \mathrm{q}_{2}=\mathrm{C}_{2}\left(\mathrm{~V}_{2}-\mathrm{V}\right)$ As $\mathrm{q}_{1}=\mathrm{q}_{2} \quad$ [capacitors are in series] $$ \begin{array}{l} \therefore C_{1}\left(V-V_{1}\right)=C_{2}\left(V_{2}-V\right) \\ V=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}} \end{array} $$

Hence, the correct answer is option: 'C'.

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