Question

In the figure shown, after the switch $'S'$ is turned from the position $'A'$ to position $'B'$ , the energy dissipated in the circuit in terms of capacitance $'C'$ and total charge $'Q'$ is:
(A) $\dfrac{3}{8}\dfrac{{{Q^2}}}{C}$
(B) $\dfrac{3}{4}\dfrac{{{Q^2}}}{C}$
(C) $\dfrac{1}{8}\dfrac{{{Q^2}}}{C}$
(D) $\dfrac{5}{8}\dfrac{{{Q^2}}}{C}$

Answer 1

Hint: Use the formula of the energy stored in the capacitor and find its answer. Use the formula of the energy liberated and substitute the value of the equivalent capacitance to find its answer. Subtract both the answer to find the value of the energy dissipated to the surrounding.

Formula used:
(1) The formula of the energy stored in the capacitor is given by
$U = \dfrac{1}{2}C{V^2}$
Where $U$ is the energy stored in the capacitor, $C$ is the capacitance and $V$ is the voltage of the capacitor.
(2) The energy dissipated in the circuit is given by
$V = \dfrac{{C{V^2}}}{{2{C_{eq}}}}$
Where $V$ is the energy dissipated to the surrounding and ${C_{eq}}$ is the equivalent capacitance of the circuit.

Complete step by step solution:
From the given circuit diagram, the capacitance of the $C$ and $3C$ are connected parallel to each other. $A$ and $B$ are two switch connections.
Using the formula of the energy stored we get
$U = \dfrac{1}{2}C{V^2}$ ……………………….(1)
Substitute that $C = \dfrac{Q}{V}$ in the above step, we get
$U = \dfrac{{{Q^2}}}{{2C}}$ ……………………..(2)
Using the formula of the energy dissipated to the surrounding, we get
$V = \dfrac{{C{V^2}}}{{2{C_{eq}}}}$ ……………………….(3)
From the circuit, the equivalent capacitance of the circuit is calculated as follows.
$\Rightarrow {C_{eq}} = 1 + 3 = 4$
Substitute this in the equation (2), we get
$\Rightarrow V = \dfrac{{C{V^2}}}{{2 \times 4}} = \dfrac{{C{V^2}}}{8}$ …………….(4)
The change in the energy is obtained by subtracting the (4) and (1),
$\Rightarrow \Delta E = \dfrac{{C{V^2}}}{2} - \dfrac{{C{V^2}}}{8} = \dfrac{{3C{V^2}}}{8}$
Substituting the equation (2) in the above step, we get
$\Rightarrow \Delta E = \dfrac{3}{8}\dfrac{{{Q^2}}}{C}$

Thus the option (A) is correct.

Note: The main work of the capacitance is to collect the energy and to store it. But there will be some loss of the energy and its mainly due to the voltage drop across the capacitor. If the voltage drop is high, then the power dissipation in the capacitor will also be high.