JEE Advanced 2017Paper 2

JEE Advanced 2017 · Paper 2 · Physics · Capacitance

Paragraph: Consider a simple R C circuit as shown in Figure 1 ....

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

. 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:

Options

A. $E_C=E_D$

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

q = CV

Energy stored in capacitor

{E}_{C} = \dfrac{1}{2}C{V}^{2}

.....(i)
Work done by battery \left(W\right) = Vq = VCV = C{V}^{2}

<br />Dissipated across resistance

{E}_{D}

= (work done by battery) – (energy store)<br />

{E}_{D} = C{V}^{2} - \dfrac{1}{2}C{V}^{2} = \dfrac{1}{2}C{V}^{2}

.....(ii)<br />From (i) and (ii)<br />

{E}_{D} = {E}_{C}