Question

Two capacitors of equal capacitance $(C_1=C_2)$ are shown in the figure. Initially, while the switch $S$ is open, one of the capacitors is uncharged and the other carries charge $Q_0$. The energy stored in the charged capacitor is $U_0$. Sometimes after the switch is closed, the capacitors $C_1$ and $C_2$ carry charges $Q_1$ and $Q_2$, respectively; the voltage across the capacitors are $V_1$ and $V_2$; and the energies stores in the capacitors are $U_1$ and $U_2$. Which of the following statements is incorrect?

• A. $Q_0=\frac{1}{2}(Q_1+Q_2)$
• B. $Q_1=Q_2$
• C. $V_1=V_2$
• D. $U_0=U_1+U_2$

Answer 1

Answer: (D)

$U_0=U_1+U_2$

In the above diagram, $C_1$ and $C_2$ are connected in parallel.

Hence, both get the same voltage. $\therefore V_1=V_2$ is correct.

We know that $Q_1=C_1{V}_1$ and $Q_2=C_2{V}_2$

$\because C_1=C_2,$ $Q_1=Q_2$

We know that energy $U_1=\frac{1}{2}{C}_1{V}_1^2$ and $U_2=\frac{1}{2}{C}_2{V}_2^2$

By closing the switch, the $C_1$ or $V_1$ doesn't get affected.

Hence $U_0=U_1$ and not $U_1+U_2$

Similarly, By closing the switch, the $Q_1$ or $V_1$ doesn't get affected.

Hence $Q_0=Q_1$

$\because Q_0=Q_1=Q_2,$ $Q_0=\frac{1}{2}(Q_1+Q_2)$

Thus D is the only wrong statement.