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Question: In the figure shown, after the switch \('S'\) is turned from the position \('A'\) to position \('B'\...

In the figure shown, after the switch S'S' is turned from the position A'A' to position B'B' , the energy dissipated in the circuit in terms of capacitance C'C' and total charge Q'Q' is:
(A) 38Q2C\dfrac{3}{8}\dfrac{{{Q^2}}}{C}
(B) 34Q2C\dfrac{3}{4}\dfrac{{{Q^2}}}{C}
(C) 18Q2C\dfrac{1}{8}\dfrac{{{Q^2}}}{C}
(D) 58Q2C\dfrac{5}{8}\dfrac{{{Q^2}}}{C}

Explanation

Solution

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=12CV2U = \dfrac{1}{2}C{V^2}
Where UU is the energy stored in the capacitor, CC is the capacitance and VV is the voltage of the capacitor.
(2) The energy dissipated in the circuit is given by
V=CV22CeqV = \dfrac{{C{V^2}}}{{2{C_{eq}}}}
Where VV is the energy dissipated to the surrounding and Ceq{C_{eq}} is the equivalent capacitance of the circuit.

Complete step by step solution:
From the given circuit diagram, the capacitance of the CC and 3C3C are connected parallel to each other. AA and BB are two switch connections.
Using the formula of the energy stored we get
U=12CV2U = \dfrac{1}{2}C{V^2} ……………………….(1)
Substitute that C=QVC = \dfrac{Q}{V} in the above step, we get
U=Q22CU = \dfrac{{{Q^2}}}{{2C}} ……………………..(2)
Using the formula of the energy dissipated to the surrounding, we get
V=CV22CeqV = \dfrac{{C{V^2}}}{{2{C_{eq}}}} ……………………….(3)
From the circuit, the equivalent capacitance of the circuit is calculated as follows.
Ceq=1+3=4\Rightarrow {C_{eq}} = 1 + 3 = 4
Substitute this in the equation (2), we get
V=CV22×4=CV28\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),
ΔE=CV22CV28=3CV28\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
ΔE=38Q2C\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.