A capacitor of capacity C_1 is charged to the | vedclass

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

Question

$U_{1}=\frac{1}{2} C_{1} V_{0}^{2}$

$C_{1} V_{0}=\left(C_{1}+C_{2}\right) V$

$V=\frac{C_{1} V_{0}}{C_{1}+C_{2}}$

$\mathrm{U}_{2}=\frac{1}{2}\

Vedclass · Accepted Answer

$\frac{{{C_1} + {C_2}}}{{{C_1}}}$

Vedclass · Answer

$U_{1}=\frac{1}{2} C_{1} V_{0}^{2}$

$C_{1} V_{0}=\left(C_{1}+C_{2}\right) V$

$V=\frac{C_{1} V_{0}}{C_{1}+C_{2}}$

$\mathrm{U}_{2}=\frac{1}{2}\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right) \frac{\mathrm{C}_{1}^{2} \mathrm{V}_{0}^{2}}{\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)^{2}}=\frac{1}{2} \frac{\mathrm{C}_{1}^{2} \mathrm{V}_{0}^{2}}{\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)}$

$\frac{U_{1}}{U_{2}}=\frac{1 / 2 C_{1} V_{0}^{2}}{1 / 2 \frac{C_{1}^{2} V_{0}^{2}}{\left(C_{1}+C_{2}\right)}}$

$\frac{U_{1}}{U_{2}}=\frac{C_{1}+C_{2}}{C_{1}}$

A capacitor of capacity $C_1$ is charged to the potential of $V_0$. After disconnecting with the battery, it is connected with a neutral capacitor of capacity $C_2$ as shown in the adjoining figure. The ratio of energy of system before and after the connection of switch $S$ will be

Similar Questions

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

A variable condenser is permanently connected to a $100$ $V$ battery. If the capacity is changed from $2\,\mu \,F$ to $10\,\mu \,F$, then change in energy is equal to

Change $Q$ on a capacitor varies with voltage $V$ as shown in the figure, where $Q$ is taken along the $X$-axis and $V$ along the $Y$-axis. The area of triangle $OAB$ represents

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

A capacitor of capacity $C$ is connected with a battery of potential $V$ in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential $V$ again, the energy given by the battery will be