Paragraph: 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 f

Question

Paragraph: 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 >> RC ). 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 3 V 0 ​ ​ and maintained for a charging time T >> RC . Then the voltage is raised to 3 2 V 0 ​ ​ without discharging the capacitor and again maintained for a time T >> RC . 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:

TestHub · 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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