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Paragraph: Consider a simple \(R C\) circuit as shown in Figure \(1\). Process...

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 \(R\). 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 \(T>>R C\). 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 2, total energy dissipated across the resistance \(E_{D}\) is:
\(E_D=\frac{1}{2} C V_0^2\)
\(E_D=3\left(\frac{1}{2} C V_0^2\right)\)
\(E_D=\frac{1}{3}\left(\frac{1}{2} C V_0^2\right)\)
\(E_D=3 C V_0^2\)
Solution:
For process (i)
Charge on capacitor =CV0 3
Energy stored in capacitor =12CV029=CV0218
Work done by battery =CV03×V3=CV029
Heat loss  =CV029-CV0218=CV0218
For process (ii)
Charge on capacitor =2CV03
Extra charge flow through battery =CV03
Work done by battery:  CV03 .2V03=2CV029
Final energy store in capacitor: 12C2V032=4CV0218
Energy store in process 2: 4CV0218-CV0218=3CV0218
Heat loss in process (ii) = work done by battery in process (ii) – energy store in capacitor process (ii)
=2CV029-3CV0218=CV0218
For process (iii)
Charge on capacitor =CV0
Extra charge flow through battery: CV0-2CV03=CV03
Work done by battery in this process: CV03 V0=CV023
Find energy store in capacitor:  12CV02
Energy stored in this process:  12CV02-4CV0218=5CV0218
Heat loss in process (iii):  CV023-5CV0218=CV0218
Now total heat loss  ED: CV0218+CV0218+CV0218=CV026
Final energy store in capacitor: 12CV02
So we can say that ED=1312CV02