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Standard XII

Physics

Charging of Capacitors

A 10 μ F capa...

Question

A $10 μ F$ capacitor charged to $20 V$ is connected across a $15 V$ battery through a switch such that positive terminal of battery is connected to negative charged plate of capacitor. When switch is closed, work done by battery of $15 V$ in long time will be

2250 μ J

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- 3000 μ J

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30000 μ J

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5250 μ J

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Solution

The correct option is D $5250 μ J$

Intial charge $Q$ on capacitor is $Q = C V = 10 \times 20 = 200 μ C$

On closing switch final charge on capacitor is $q = C ε = 10 \times 15 μ C = 150 μ C$
So charge flown through battery is $Δ q = q + Q = 350 μ C$
So work of battery is $W = ε Δ q$
$W = 15 \times 350 μ J = 5250 μ J$

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A 10 μF capacitor charged to 20 V is connected across a 15 V battery through a switch such that positive terminal of battery is connected to negative charged plate of capacitor. When switch is closed, work done by battery of 15 V in long time will be

The correct option is D 5250 μJ Intial charge Q on capacitor is Q=CV=10×20=200 μC On closing switch final charge on capacitor is q=Cε=10×15 μC=150 μC So charge flown through battery is Δq=q+Q=350 μC So work of battery is W=εΔq W=15×350 μJ=5250 μJ

A $10 μ F$ capacitor charged to $20 V$ is connected across a $15 V$ battery through a switch such that positive terminal of battery is connected to negative charged plate of capacitor. When switch is closed, work done by battery of $15 V$ in long time will be