How does a capacitor store energy ? And obtain the formula for the energy stored in the capacitor ?
How does a capacitor store energy ? And obtain the formula for the energy stored in the capacitor ?
Take uncharged conductors $1$ and $2$ as shown in figure.
Imagine a process of transferring charge from conductor $2$ to conductor $1$ bit by bit so that at the ends, conductor $1$ gets charge.
In transferring positive charge from conductor$2$ to conductor 1 work will be done externally, because at any stage conductor $1$ is at a higher potential than conductor $2$ .
To calculate the total work by in a small step involving transfer of an infinitesimally amount of charge.
Consider the situation when the conductors $1$ and $2$ have charges $Q'$ and $- Q'$ respectively.
The potential difference $V^{\prime}$ between conductors $1$ to $2$ is $V^{\prime}=\frac{Q^{\prime}}{C}$ where $C$ is the capacitance of the system.
A small charge $\delta Q^{\prime}$ is transferred from conductor $2$ to $1$ , then work done, $\delta \mathrm{W}=\mathrm{V}^{\prime} \delta Q^{\prime}$
$\therefore \delta \mathrm{W}=\frac{\mathrm{Q}^{\prime} \delta \mathrm{Q}^{\prime}}{\mathrm{C}}$
$\ldots$ $(1)$
Total work done to bringing charge $Q$ from conductor 2 to 1 is obtain by integration, $\mathrm{W}=\int d \mathrm{~W}$
$\therefore \mathrm{W}=\int_{0}^{\mathrm{Q}} \frac{\mathrm{Q}^{\prime}}{\mathrm{C}} \cdot \delta \mathrm{Q}^{\prime} \quad \therefore \mathrm{W}=\frac{1}{\mathrm{C}} \int_{0}^{\mathrm{Q}} \mathrm{Q}^{\prime} \delta \mathrm{Q}^{\prime}=\frac{1}{\mathrm{C}}\left[\frac{\left(\mathrm{Q}^{\prime}\right)^{2}}{2}\right]_{0}^{\mathrm{Q}}$
$\frac{1}{\mathrm{C}}\left[\frac{\mathrm{Q}^{2}}{2}\right] \therefore \mathrm{W}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}$
Similar Questions
Consider a simple $RC$ circuit as shown in Figure $1$.
Process $1$: In the circuit the switch $S$ is closed at $t=0$ and the capacitor is fully charged to voltage $V_0$ (i.e. charging continues for time $T \gg R C$ ). In the process some dissipation ( $E_D$ ) occurs across the resistance $R$. The amount of energy finally stored in the fully charged capacitor is $EC$.
Process $2$: In a different process the voltage is first set to $\frac{V_0}{3}$ and maintained for a charging time $T \gg R C$. Then the voltage is raised to $\frac{2 \mathrm{~V}_0}{3}$ without discharging the capacitor and again maintained for time $\mathrm{T} \gg \mathrm{RC}$. The process is repeated one more time by raising the voltage to $V_0$ and the capacitor is charged to the same final
take $\mathrm{V}_0$ as voltage
These two processes are depicted in Figure $2$.
($1$) In Process $1$, the energy stored in the capacitor $E_C$ and heat dissipated across resistance $E_D$ are released by:
$[A]$ $E_C=E_D$ $[B]$ $E_C=E_D \ln 2$ $[C]$ $\mathrm{E}_{\mathrm{C}}=\frac{1}{2} \mathrm{E}_{\mathrm{D}}$ $[D]$ $E_C=2 E_D$
($2$) In Process $2$, total energy dissipated across the resistance $E_D$ is:
$[A]$ $\mathrm{E}_{\mathrm{D}}=\frac{1}{2} \mathrm{CV}_0^2$ $[B]$ $\mathrm{E}_{\mathrm{D}}=3\left(\frac{1}{2} \mathrm{CV}_0^2\right)$ $[C]$ $\mathrm{E}_{\mathrm{D}}=\frac{1}{3}\left(\frac{1}{2} \mathrm{CV}_0^2\right)$ $[D]$ $\mathrm{E}_{\mathrm{D}}=3 \mathrm{CV}_0^2$
Given the answer quetion ($1$) and ($2$)
Consider a simple $RC$ circuit as shown in Figure $1$.
Process $1$: In the circuit the switch $S$ is closed at $t=0$ and the capacitor is fully charged to voltage $V_0$ (i.e. charging continues for time $T \gg R C$ ). In the process some dissipation ( $E_D$ ) occurs across the resistance $R$. The amount of energy finally stored in the fully charged capacitor is $EC$.
Process $2$: In a different process the voltage is first set to $\frac{V_0}{3}$ and maintained for a charging time $T \gg R C$. Then the voltage is raised to $\frac{2 \mathrm{~V}_0}{3}$ without discharging the capacitor and again maintained for time $\mathrm{T} \gg \mathrm{RC}$. The process is repeated one more time by raising the voltage to $V_0$ and the capacitor is charged to the same final
take $\mathrm{V}_0$ as voltage
These two processes are depicted in Figure $2$.
($1$) In Process $1$, the energy stored in the capacitor $E_C$ and heat dissipated across resistance $E_D$ are released by:
$[A]$ $E_C=E_D$ $[B]$ $E_C=E_D \ln 2$ $[C]$ $\mathrm{E}_{\mathrm{C}}=\frac{1}{2} \mathrm{E}_{\mathrm{D}}$ $[D]$ $E_C=2 E_D$
($2$) In Process $2$, total energy dissipated across the resistance $E_D$ is:
$[A]$ $\mathrm{E}_{\mathrm{D}}=\frac{1}{2} \mathrm{CV}_0^2$ $[B]$ $\mathrm{E}_{\mathrm{D}}=3\left(\frac{1}{2} \mathrm{CV}_0^2\right)$ $[C]$ $\mathrm{E}_{\mathrm{D}}=\frac{1}{3}\left(\frac{1}{2} \mathrm{CV}_0^2\right)$ $[D]$ $\mathrm{E}_{\mathrm{D}}=3 \mathrm{CV}_0^2$
Given the answer quetion ($1$) and ($2$)
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