A 2 mu F capacitor is charged as shown in fi | vedclass

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

Question

Initial energy stored in capacitor $2 \mu F : \quad U _{ i }=\frac{1}{2} 2(V)^2= V ^2$

Final voltage after switch 2 is on: $V _{ f }=\frac{ C _1 V_

Vedclass · Accepted Answer

$80 \%$

Vedclass · Answer

Initial energy stored in capacitor $2 \mu F : \quad U _{ i }=\frac{1}{2} 2(V)^2= V ^2$

Final voltage after switch 2 is on: $V _{ f }=\frac{ C _1 V_1}{ C _1+ C _2}=\frac{2 V}{1 O }=0.2 V$

Final energy in both the capacitors,

$U _{ f }=\frac{1}{2}\left( C _1+ C _2ight) V _{ f }^2=\frac{1}{2} 	imes 10 	imes\left(\frac{2 V}{10}ight)^2=0.2 V^2$

Therefore, energy dissipated $=\frac{ V ^2-0.2 V^2}{V^2} 	imes 100=80 \%$

A $2 \ \mu F$ capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position $2$ is

Similar Questions

A $4 \;\mu\, F$ capacitor is charged by a $200\; V$ supply. It is then disconnected from the supply, and is connected to another uncharged $2 \;\mu\, F$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Two spherical conductors each of capacity $C$ are charged to potentials $V$ and $ - V$. These are then connected by means of a fine wire. The loss of energy will be

Obtain the expression for the energy stored per unit volume in a charged capacitor.

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

If $E$ is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to