Gujarati
Hindi
2. Electric Potential and Capacitance
normal

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  ($2$)

A

$B,C$

B

$B,D$

C

$A,C$

D

$A,D$

(IIT-2017)

Solution

($1$) $\mathrm{W}_{\mathrm{b}}=\mathrm{CV}_0 \times \mathrm{V}_0$

$\Delta \mathrm{U}=\frac{1}{2} \mathrm{CV}_0^2=\mathrm{E}_{\mathrm{C}}$

Heat dissipated $=\mathrm{W}_{\mathrm{b}}-\mathrm{E}_{\mathrm{C}}$ or $\mathrm{E}_{\mathrm{C}}=\mathrm{E}_{\mathrm{D}}=\frac{1}{2} \mathrm{CV}_0^2$

($2$) If capacitor is charged from $V_i$ to $V_f$; heat dissipated is

$\mathrm{H}=\mathrm{W}_{\text {battery }}-\Delta \mathrm{U}$

$=\mathrm{C}\left(\mathrm{V}_{\mathrm{f}}-\mathrm{V}_{\mathrm{i}}\right) \mathrm{V}_{\mathrm{f}}-\left[\frac{1}{2} \mathrm{C}\left(\mathrm{V}_{\mathrm{f}}^2-\mathrm{V}_{\mathrm{i}}^2\right)\right]$

$=\frac{1}{2} \mathrm{C}\left(\mathrm{V}_{\mathrm{f}}-\mathrm{V}_{\mathrm{i}}\right)^2$

Total heat dissipated

$=\frac{1}{2} \mathrm{C}\left\{\left(\frac{\mathrm{V}_0}{3}-0\right)^2+\left(\frac{2 \mathrm{~V}_0}{3}-\frac{\mathrm{V}_0}{3}\right)^2+\left(\mathrm{V}_0-\frac{2 \mathrm{~V}_0}{3}\right)^2\right\}$

$=\frac{1}{6} \mathrm{CV} \mathrm{V}_0^2$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.