Question 692: Electrostatics & Capacitance

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 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 $${{{V_0}} \over 3}$$ and maintained for a charging time T >> RC. Then, the voltage is raised to $${{2{V_0}} \over 3}$$ without discharging the capacitor and again maintained for a time T >> RC. The process is repeated one more time by raising the voltage to 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.

Accepted Answer

Energy supplied to the circuit is CV$$_0^2$$
Energy stored in capacitor, E
C
= $${1\over 2}$$CV$$_0^2$$
Therefore,
Energy dissipated E
0
= Energy supplied $$-$$ Energy stored
= CV$$_0^2$$ $$-$$ $${1\over 2}$$CV$$_0^2$$ = $${1\over 2}$$CV$$_0^2$$
$$\Rightarrow$$ E
C
= E
D

Answer

E
C
= E
D
ln2

Answer

E
C
= E
D

Answer

E
C
= 2E
D

Answer

E
C
= $${1 \over 2}$$E
D

Electrostatics & Capacitance

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