PARAGRAPH 1 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 gg 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 EC. Process 2: In a different process the voltage is first set to ( V 0/3) and maintained for a charging time T gg RC. Then the voltage is raised to (2 V 0/3) without discharging the capacitor and again maintained for time T gg 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 o as in Process 1 . These two processes are depicted in Figure 2 . In Process 1, the energy stored in the capacitor EC and heat dissipated across resistance ED are released by:

Question

PARAGRAPH 1 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 gg 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 EC. Process 2: In a different process the voltage is first set to ( V 0/3) and maintained for a charging time T gg RC. Then the voltage is raised to (2 V 0/3) without discharging the capacitor and again maintained for time T gg 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 o as in Process 1 . These two processes are depicted in Figure 2 . In Process 1, the energy stored in the capacitor EC and heat dissipated across resistance ED are released by: (A) 𝐸C = 𝐸D (B) 𝐸C = 𝐸D ln 2 (C) EC = (1/2)ED (D) 𝐸C = 2𝐸D

Tardigrade Admin · Accepted Answer

W b = CV 0 × V 0 Δ U =(1/2) CV 02= E C Heat dissipated = W b - E C or E C = E D =(1/2) CV 02

$E_{C} = E_{D}$

$EC = E_{D}$ ln 2

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

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

$W_{b} = C V_{0} \times V_{0}$
$Δ U = \frac{1}{2} C V_{0}^{2} = E_{C}$
Heat dissipated $= W_{b} - E_{C}$
or $E_{C} = E_{D} = \frac{1}{2} C V_{0}^{2}$

EC​=ED​

EC=ED​ ln 2

EC​=21​ED​

EC​=2ED​

Wb​=CV0​×V0​ ΔU=21​CV02​=EC​ Heat dissipated =Wb​−EC​ or EC​=ED​=21​CV02​

$E_{C} = E_{D}$

$EC = E_{D}$ ln 2

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

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

$W_{b} = C V_{0} \times V_{0}$
$Δ U = \frac{1}{2} C V_{0}^{2} = E_{C}$
Heat dissipated $= W_{b} - E_{C}$
or $E_{C} = E_{D} = \frac{1}{2} C V_{0}^{2}$