First slide

Capacitance

Question

A capacitor of capacitance 160 $μ$ F is charged to a potential difference of 200 V and then connected across a discharge tube which conducts until the potential difference across it has fallen to 100 V. The energy dissipated in the tube is

Moderate

A

6.4 J
B

4.8 J
C

3.2 J
D

2.4 J

Solution

$\begin{array}{l} E = \frac{1}{2} C (V^{2} - V^{' 2}) \\ = \frac{1}{2} \times (160 \times 10^{- 6}) [(200)^{2} - (100)^{2}] \\ = \frac{1}{2} \times (160 \times 10^{- 6}) \times (3 \times 10^{4}) = 2 .4 J \end{array}$

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