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 if fully charged to voltage V₀ (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 $\frac{V_{0}}{3}$ and maintained for a charging time T >>RC. Then the voltage is raised to $\frac{2 V_{0}}{3}$ without discharging the capacitor and again maintained for a time T >> RC. The process is repeated one more time by raising the voltage to V₀ and the capacitor is charged to the same final voltage V₀ as in Process 1.

These two processes are depicted in figure 2.

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Linked Question 1

In Process 1, the energy stored in the capacitor E_C and heat dissipated across resistance E_D are related by :

E_C = E_D

E_C = 2E_D

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

E_C = E_D ln 2

$E = E_{D} = \frac{1}{2} {CV}_{0}^{2}$

Linked Question 2

In Process 2, total energy dissipated across the resistance E_D is:

$E_{D} = 3 (\frac{1}{2} {CV}_{0}^{2})$

$E_{D} = \frac{1}{2} {CV}_{0}^{2}$

$E_{D} = 3 {CV}_{0}^{2}$

$E_{D} = \frac{1}{3} (\frac{1}{2} {CV}_{0}^{2})$

E_D = Work done by battery – energy stored in capacitor

= q₁ $\frac{V_{0}}{3}$ + q₃ $\frac{2 V_{0}}{3}$ + q₃V₀ – $\frac{1}{2}$ CV₀²

q = $\frac{{CV}_{0}}{3}$ q₂ = $\frac{{CV}_{0}}{3}$ q₃ = $\frac{{CV}_{0}}{3}$

ED = $\frac{{CV}_{0}^{2}}{9} + \frac{2 {CV}_{0}^{2}}{9} + \frac{3 {CV}_{0}^{2}}{9} - \frac{1}{2}$ CV₀²

ED = $\frac{1}{6}$ CV₀²

^{$E_{D} = \frac{1}{3} (\frac{1}{2} {CV}_{0}^{2})$}