A capacitor having a capacity of 2.0 ; mu F i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.Given:Capacitance $C = 2.0 \; \mu F = 2.0 	imes 10^{-6} \; F$Pot

Vedclass · Accepted Answer

$4 	imes 10^{-2} \; J$

Vedclass · Answer

(C) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.Given:Capacitance $C = 2.0 \; \mu F = 2.0 	imes 10^{-6} \; F$Potential difference $V = 200 \; V$Substituting the values into the formula:$U = \frac{1}{2} 	imes (2.0 	imes 10^{-6} \; F) 	imes (200 \; V)^2$$U = 1.0 	imes 10^{-6} 	imes 40000$$U = 4 	imes 10^{-2} \; J$When the capacitor is discharged through a resistance wire,the entire stored energy is dissipated as heat. Therefore,the heat produced is $4 	imes 10^{-2} \; J$.

$A$ capacitor having a capacity of $2.0 \; \mu F$ is charged to $200 \; V$ and then the plates of the capacitor are connected to a resistance wire. The heat produced in joules will be:

Similar Questions

The work done in placing a charge of $8 \times 10^{-18} \ C$ on a capacitor of $800 \ \mu F$ is:

How does a capacitor store energy? Obtain the formula for the energy stored in a capacitor.

The total energy stored in the capacitor system shown in the figure will be (in $\mu J$)

If the energy density in a parallel plate capacitor is given as $1.8 \times 10^{-9} \ J/m^3$,then the value of the electric field in the region between the plates is ....... $NC^{-1}$. (Given: $\epsilon_0 = 9 \times 10^{-12} \ C^2/N \cdot m^2$)

The capacity of a capacitor is $4 \times 10^{-6} \, F$ and its potential is $100 \, V$. The energy released on discharging it fully will be ....... $J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module