In Process 1, the energy stored in the capacitor and heat dissipated across resistance are related by: [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 1, the energy stored in the capacitor $E_{C}$ and heat dissipated across resistance $E_{D}$ are related by: [2017]

Smart Achievers · Accepted Answer

(1)

In process 1, work done by the battery, $W = q \times V$ $= C V_{0} \times V_{0}$ $= C V_{0}^{2}$

Energy stored in the battery, $E_{C} = \frac{1}{2} C V_{0}^{2}$

Heat dissipated across the resistance,

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

Therefore, $E_{C} = E_{D}$

Solution

1 $E_{C} = E_{D}$

2 $E_{C} = E_{D} \ln 2$

3 $E_{C} = \frac{1}{2} E_{D}$

4 $E_{C} = 2 E_{D}$

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Solution

1 EC=ED

2 EC=EDln2

3 EC=12ED

4 EC=2ED

Ans. (1) In process 1, work done by the battery, W=q×V=CV0×V0=CV02 Energy stored in the battery, EC=12CV02 Heat dissipated across the resistance, ED=W-EC=CV02-12CV02=12CV02 Therefore, EC=ED

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1 $E_{C} = E_{D}$

2 $E_{C} = E_{D} \ln 2$

3 $E_{C} = \frac{1}{2} E_{D}$

4 $E_{C} = 2 E_{D}$