(B) The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.Given: $C = 40\ \mu F = 40 \times 10^{-6}\ F$,$V = 3000\ V$,and $t = 2\ ms = 2

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(B) The energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$.Given: $C = 40\ \mu F = 40 \times 10^{-6}\ F$,$V = 3000\ V$,and $t = 2\ ms = 2 \times 10^{-3}\ s$.Substituting the values:$U = \frac{1}{2} \times (40 \times 10^{-6}) \times (3000)^2 = 20 \times 10^{-6} \times 9 \times 10^6 = 180\ J$.The power delivered is $P = \frac{U}{t} = \frac{180\ J}{2 \times 10^{-3}\ s} = 90 \times 10^3\ W = 90\ kW$.

Class 12 Physics Electric Potential and Capacitance Energy Stored in a Capacitor

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In a defibrillator,a $40\ \mu F$ capacitor is charged to $3000\ V$. If the stored energy of the capacitor is discharged through the patient during a pulse of duration $2\ ms$,what is the power delivered to the patient in $kW$?

A
$45$
B
$90$
C
$180$
D
$360$

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Class 12

Physics Questions

Chapter

Electric Potential and Capacitance

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Energy Stored in a Capacitor

The amount of work done in increasing the voltage across the plates of a capacitor from $5 \ V$ to $10 \ V$ is $W$. The work done in increasing it from $10 \ V$ to $15 \ V$ will be (nearly):

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If the potential difference across a capacitor is increased from $5 \,V$ to $15 \,V$, then the ratio of final energy to initial energy stored in the capacitor is

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The charge $q$ on a capacitor varies with voltage $V$ as shown in the figure. The area of the triangle $AOB$ represents:

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$A$ capacitor of $6 \ \mu F$ is charged from $10 \ V$ to $20 \ V$. The increase in its stored energy is:

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If the distance between the plates of a capacitor with capacitance $C$ and charge $Q$ is doubled,the work done is:

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In a defibrillator,a $40\ \mu F$ capacitor is charged to $3000\ V$. If the stored energy of the capacitor is discharged through the patient during a pulse of duration $2\ ms$,what is the power delivered to the patient in $kW$?

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The amount of work done in increasing the voltage across the plates of a capacitor from $5 \ V$ to $10 \ V$ is $W$. The work done in increasing it from $10 \ V$ to $15 \ V$ will be (nearly):

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