Paragraph: Consider a simple $R C$ 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 $TR C$). In the process some dissipation $\left(E_{D}\right)$ occurs across the…

Question

Paragraph: Consider a simple $R C$ 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 $T>>R C$). In the process some dissipation $\left(E_{D}\right)$ 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 $2 .$ Question: In Process 1, the energy stored in the capacitor $E_{C}$ and heat dissipated across resistance $E_{D}$ are related by:

Aicharya · Accepted Answer

When switch is closed for a very long time capacitor will get fully charged and charge on capacitor will be

q = C V

Energy stored in capacitor

E_{C} = \frac{1}{2} C V^{2}

.....(i)
Work done by battery

(W) = V q = V C V = C V^{2}

Dissipated across resistance

E_{D}

= (work done by battery) – (energy store)

E_{D} = C V^{2} - \frac{1}{2} C V^{2} = \frac{1}{2} C V^{2}

.....(ii)
From (i) and (ii)

E_{D} = E_{C}

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Paragraph: Consider a simple $R C$ circuit as shown in Figure $1$. Process 1: In the circuit the switch $S$…

Solution