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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 1 , the energy stored in the capacitor \(E_{C}\) and heat dissipated across resistance \(E_{D}\) are related by:
\(E_C=E_D\)
\(E_C=E_D \ln 2\)
\(E_C=\frac{1}{2} E_D\)
\(E_C=2 E_D\)
Solution:

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 EC=12CV2      .....(i)
Work done by battery W=Vq=VCV=CV2
Dissipated across resistance ED = (work done by battery) – (energy store)
ED=CV2-12CV2=12CV2        .....(ii)
From (i) and (ii)
ED= EC