Capacity of a capacitor is 48 mu F. When it i | vedclass

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

Question

(A) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.Given: $C = 48 \mu F = 48 	imes 10^{-6} \ F$,$q_1 = 0.1 \ C$,$q_2 = 0.5 \ C$.Th

Vedclass · Accepted Answer

$2500 \ J$

Vedclass · Answer

(A) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.Given: $C = 48 \mu F = 48 	imes 10^{-6} \ F$,$q_1 = 0.1 \ C$,$q_2 = 0.5 \ C$.The change in energy $\Delta U$ is given by:$\Delta U = U_2 - U_1 = \frac{q_2^2}{2C} - \frac{q_1^2}{2C} = \frac{1}{2C} (q_2^2 - q_1^2)$Substituting the values:$\Delta U = \frac{1}{2 	imes 48 	imes 10^{-6}} ((0.5)^2 - (0.1)^2)$$\Delta U = \frac{1}{96 	imes 10^{-6}} (0.25 - 0.01)$$\Delta U = \frac{0.24}{96 	imes 10^{-6}}$$\Delta U = \frac{0.24 	imes 10^6}{96} = \frac{240000}{96} = 2500 \ J$.

Capacity of a capacitor is $48 \mu F$. When it is charged from $0.1 C$ to $0.5 C$,the change in the energy stored is:

Similar Questions

$A$ capacitor of capacity $C$ has charge $Q$ and stored energy is $W$. If the charge is increased to $2Q$,the stored energy will be

$A$ $12 \text{ pF}$ capacitor is connected to a $50 \text{ V}$ battery. The electrostatic energy stored in the capacitor will be . . . . . . $\text{J}$.

$A$ parallel plate capacitor having a plate separation of $2\, mm$ is charged by connecting it to a $300\, V$ supply. The energy density is.....$J/m^3$

$A$ parallel plate air capacitor has a uniform electric field $E$ in the space between the plates. The area of each plate is $A$ and the distance between the plates is $d$. The energy stored in the capacitor is $[\varepsilon_0 = \text{permittivity of free space}]$

The charge $q$ on a capacitor varies with voltage $V$ as shown in the graph. The area of $\Delta OPM$ represents:

Vedclass Test Series

Exam Paper Generator

Online Exam Module