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2. Electric Potential and Capacitance
easy
$6\ \mu F$ ના કેપેસીટરને $10\, volts$ થી $20 \,volts$ વિદ્યુતભારીત કરેલ છે તો ઉર્જામાં થતો વધારો.....
A
$18 \times {10^{ - 4}}J$
B
$9 \times {10^{ - 4}}J$
C
$4.5 \times {10^{ - 4}}J$
D
$9 \times {10^{ - 6}}J$
Solution
$\Delta E = E_{final} – E_{initial} = \frac{1}{2}C\,(V_{Final}{^2} – V_{Initial}{^2})$
$ = \frac{1}{2} \times 6 \times ({20^2} – {10^2}) \times {10^{ – 6}}\, = 3 \times (400 – 100) \times {10^{ – 6}} = 3 \times 300 \times {10^{ – 6}} = 9 \times {10^{ – 4}}\ J$
Standard 12
Physics
