In Process 2, total energy dissipated across the resistance ED is: [2017]

Question

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 the voltage $V_{0}$ (i.e., charging continues for time $T ≫ R C$ ). In the process some dissipation $(E_{D})$ occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is $E_{C}$ .

Process 2: In a different process the voltage is first set to $\frac{V_{0}}{3}$ and maintained for a charging time $T ≫ R C$ . Then the voltage is raised to $\frac{2 V_{0}}{3}$ without discharging the capacitor and again maintained for a time $T ≫ R C$ . The process is repeated one more time by raising the voltage to $V_{0}$ and the capacitor is charged to the same final voltage $V_{0}$ as in Process 1.

These two processes are depicted in Figure 2.

Q. In Process 2, total energy dissipated across the resistance $E_{D}$ is: [2017]

Smart Achievers · Accepted Answer

(3)

Let $V_{i}$ and $V_{f}$ be the initial and final voltages in each step of process 2. Then,

$T ≫ R C$

$= C (V_{f} - V_{i}) V_{f} - \frac{1}{2} C {(V_{f} - V_{i})}^{2}$

$= \frac{1}{2} C {(V_{f} - V_{i})}^{2}$

Therefore, the total energy dissipated across the resistance is

$E_{D} = \frac{1}{2} C [{(\frac{V_{0}}{3} - 0)}^{2} + {(\frac{2 V_{0}}{3} - \frac{V_{0}}{3})}^{2} + {(V_{0} - \frac{2 V_{0}}{3})}^{2}]$

$= \frac{1}{2} C [\frac{V_{0}^{2}}{9} + \frac{V_{0}^{2}}{9} + \frac{V_{0}^{2}}{9}]$

$= \frac{1}{2} C (\frac{V_{0}^{2}}{3})$

$= \frac{1}{6} C V_{0}^{2}$

Hence, $E_{D} = \frac{1}{6} C V_{0}^{2}$

Solution

1 $E_{D} = \frac{1}{2} C V_{0}^{2}$

2 $E_{D} = 3 (\frac{1}{2} C V_{0}^{2})$

3 $E_{D} = \frac{1}{3} (\frac{1}{2} C V_{0}^{2})$

4 $E_{D} = 3 C V_{0}^{2}$

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Solution

1 ED=12CV02

2 ED=3(12CV02)

3 ED=13(12CV02)

4 ED=3CV02

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1 $E_{D} = \frac{1}{2} C V_{0}^{2}$

2 $E_{D} = 3 (\frac{1}{2} C V_{0}^{2})$

3 $E_{D} = \frac{1}{3} (\frac{1}{2} C V_{0}^{2})$

4 $E_{D} = 3 C V_{0}^{2}$