Two capacitors each of 1,μ F capacitance are connected in parallel and are then charged by 200volts

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2. Electric Potential and Capacitance

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Two capacitors each of $1\,\mu F$ capacitance are connected in parallel and are then charged by $200\;volts$ $d.c.$ supply. The total energy of their charges (in $joules$) is

A

$0.01$

B

$0.02$

C

$0.04$

D

$0.06$

Solution

(c) $U = \frac{1}{2}C{V^2} = \frac{1}{2} \times 2 \times {(200)^2} \times {10^{ – 6}} = 0.04\,J$

Standard 12

Physics

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