A 6,mu F capacitor is charged from 10,V to 20 | vedclass

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

Question

(B) The energy stored in a capacitor is given by $E = \frac{1}{2}CV^2$.The increase in energy $\Delta E$ is given by $\Delta E = E_{Final} - E_{Initia

Vedclass · Accepted Answer

$9 	imes 10^{-4}\,J$

Vedclass · Answer

(B) The energy stored in a capacitor is given by $E = \frac{1}{2}CV^2$.The increase in energy $\Delta E$ is given by $\Delta E = E_{Final} - E_{Initial} = \frac{1}{2}C(V_{Final}^2 - V_{Initial}^2)$.Given: $C = 6\,\mu F = 6 	imes 10^{-6}\,F$,$V_{Initial} = 10\,V$,$V_{Final} = 20\,V$.Substituting the values:$\Delta E = \frac{1}{2} 	imes (6 	imes 10^{-6}) 	imes (20^2 - 10^2)$$\Delta E = 3 	imes 10^{-6} 	imes (400 - 100)$$\Delta E = 3 	imes 10^{-6} 	imes 300$$\Delta E = 900 	imes 10^{-6}\,J = 9 	imes 10^{-4}\,J$.

$A$ $6\,\mu F$ capacitor is charged from $10\,V$ to $20\,V$. The increase in energy will be:

Similar Questions

$A$ parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and the area of each plate is $A,$ the energy stored in the capacitor is

$A$ capacitor has a capacity of $2\,\mu F$ and is charged to a voltage of $50\,V$. The energy stored is:

The energy stored in a capacitor of capacity $C$ which has been raised to a potential $V$ is given by

$A$ capacitor is used to store $24 \, \text{watt-hour}$ of energy at $1200 \, \text{V}$. What should be the capacitance of the capacitor?

Energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential difference $V$ is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module