First slide

Capacitance

Question

Two identical capacitors, have the same capacitance C. One of them is charged to potential V₁and the other to V₂. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

Easy

A

$\frac{1}{4} C (V_{1}^{2} - V_{2}^{2})$
B

$\frac{1}{4} C (V_{1}^{2} + V_{2}^{2})$
C

$\frac{1}{4} C {(V_{1} - V_{2})}^{2}$
D

$\frac{1}{4} C {(V_{1} + V_{2})}^{2}$

Solution

Common potential $V = \frac{{CV}_{1} + {CV}_{2}}{C + C} = (\frac{V_{1} + V_{2}}{2})$
$\begin{array}{l} Loss = (\frac{1}{2} {CV}_{1}^{2} + \frac{1}{2} {CV}_{2}^{2}) - \frac{1}{2} (C + C) V^{2} \\ = \frac{1}{4} C {(V_{1} - V_{2})}^{2} \end{array}$

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