First slide

Capacitance

Question

A 4 $μ$ F capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 2 $μ$ F capacitor. The energy lost in the process is

Moderate

A

$3 \times 10^{- 2} J$
B

$3 \times 10^{2} J$
C

$2 \cdot 67 \times 10^{- 2} J$
D

$2 \cdot 8 \times 10^{2} J$

Solution

The initial energy E₁ is given by
$\begin{array}{l} E_{1} = \frac{1}{2} C_{1} V_{1}^{2} = \frac{1}{2} \times (4 \times 10^{- 6}) (200)^{2} \\ = 8 \times 10^{- 2} J \end{array}$
The common potential V is given by
$V = \frac{q_{1} + q_{2}}{C_{1} + C_{2}} = \frac{4 \times 200 + 0}{4 + 2} = \frac{400}{3} V$
The final energy E₂ is given by
$\begin{array}{l} E_{2} = \frac{1}{2} (C_{1} + C_{2}) V^{2} \\ = \frac{1}{2} (6 \times 10^{- 6}) (400 / 3)^{2} = 5 \cdot 33 \times 10^{- 2} J \end{array}$
Loss of energy $= E_{1} - E_{2}$
$= 8 \times 10^{- 2} - 5 \cdot 33 \times 10^{- 2} = 2 \cdot 67 \times 10^{- 2} J$

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