Understanding the Physics

In any process of charging a capacitor through a resistor, the energy supplied by the voltage source is distributed between the energy stored in the capacitor and the energy dissipated as heat in the resistor. According to the law of conservation of energy: Energy supplied by source ($W_{source}$) = Increase in energy stored in capacitor ($\Delta E_C$) + Energy dissipated in resistor ($E_D$). Therefore, $E_D = W_{source} - \Delta E_C$.

When a capacitor is charged from an initial voltage $V_i$ to a final voltage $V_f$ by a source of constant voltage $V_s$, the following calculations apply:

1. The charge that flows from the source is $\Delta Q = C(V_f - V_i)$.
2. The energy supplied by the source is $W_{source} = \Delta Q \cdot V_s = C(V_f - V_i)V_s$.
3. The increase in energy stored in the capacitor is $\Delta E_C = \frac{1}{2}CV_f^2 - \frac{1}{2}CV_i^2$.

We will apply these principles to each stage of Process 2.

Step-by-Step Calculation for Process 2

Stage 1: Charging from 0 to $V_0/3$

• Source voltage, $V_{s1} = V_0/3$.
• Initial capacitor voltage, $V_{i1} = 0$.
• Final capacitor voltage, $V_{f1} = V_0/3$.

1. Charge supplied: $\Delta Q_1 = C(V_{f1} - V_{i1}) = C(\frac{V_0}{3} - 0) = \frac{CV_0}{3}$.
2. Energy supplied by source: $W_1 = \Delta Q_1 \cdot V_{s1} = (\frac{CV_0}{3})(\frac{V_0}{3}) = \frac{1}{9}CV_0^2$.
3. Increase in stored energy: $\Delta E_{C1} = \frac{1}{2}CV_{f1}^2 - \frac{1}{2}CV_{i1}^2 = \frac{1}{2}C(\frac{V_0}{3})^2 - 0 = \frac{1}{18}CV_0^2$.
4. Energy dissipated: $E_{D1} = W_1 - \Delta E_{C1} = \frac{1}{9}CV_0^2 - \frac{1}{18}CV_0^2 = \frac{1}{18}CV_0^2$.

Stage 2: Charging from $V_0/3$ to $2V_0/3$

• Source voltage, $V_{s2} = 2V_0/3$.
• Initial capacitor voltage, $V_{i2} = V_0/3$.
• Final capacitor voltage, $V_{f2} = 2V_0/3$.

1. Charge supplied: $\Delta Q_2 = C(V_{f2} - V_{i2}) = C(\frac{2V_0}{3} - \frac{V_0}{3}) = \frac{CV_0}{3}$.
2. Energy supplied by source: $W_2 = \Delta Q_2 \cdot V_{s2} = (\frac{CV_0}{3})(\frac{2V_0}{3}) = \frac{2}{9}CV_0^2$.
3. Increase in stored energy: $\Delta E_{C2} = \frac{1}{2}C(V_{f2}^2 - V_{i2}^2) = \frac{1}{2}C[(\frac{2V_0}{3})^2 - (\frac{V_0}{3})^2] = \frac{1}{2}C(\frac{4V_0^2}{9} - \frac{V_0^2}{9}) = \frac{1}{2}C(\frac{3V_0^2}{9}) = \frac{1}{6}CV_0^2$.
4. Energy dissipated: $E_{D2} = W_2 - \Delta E_{C2} = \frac{2}{9}CV_0^2 - \frac{1}{6}CV_0^2 = \frac{4-3}{18}CV_0^2 = \frac{1}{18}CV_0^2$.

Stage 3: Charging from $2V_0/3$ to $V_0$

• Source voltage, $V_{s3} = V_0$.
• Initial capacitor voltage, $V_{i3} = 2V_0/3$.
• Final capacitor voltage, $V_{f3} = V_0$.

1. Charge supplied: $\Delta Q_3 = C(V_{f3} - V_{i3}) = C(V_0 - \frac{2V_0}{3}) = \frac{CV_0}{3}$.
2. Energy supplied by source: $W_3 = \Delta Q_3 \cdot V_{s3} = (\frac{CV_0}{3})(V_0) = \frac{1}{3}CV_0^2$.
3. Increase in stored energy: $\Delta E_{C3} = \frac{1}{2}C(V_{f3}^2 - V_{i3}^2) = \frac{1}{2}C[V_0^2 - (\frac{2V_0}{3})^2] = \frac{1}{2}C(V_0^2 - \frac{4V_0^2}{9}) = \frac{1}{2}C(\frac{5V_0^2}{9}) = \frac{5}{18}CV_0^2$.
4. Energy dissipated: $E_{D3} = W_3 - \Delta E_{C3} = \frac{1}{3}CV_0^2 - \frac{5}{18}CV_0^2 = \frac{6-5}{18}CV_0^2 = \frac{1}{18}CV_0^2$.

Total Energy Dissipated in Process 2

The total energy dissipated, $E_D$, is the sum of the energies dissipated in each stage: $E_D = E_{D1} + E_{D2} + E_{D3}$ $E_D = \frac{1}{18}CV_0^2 + \frac{1}{18}CV_0^2 + \frac{1}{18}CV_0^2 = 3 \times \frac{1}{18}CV_0^2 = \frac{1}{6}CV_0^2$

Comparing with Options

We need to express the result in the format given in the options. The options are in terms of $\frac{1}{2}CV_0^2$. Let's analyze Option A: ${E_D} = {1 \over 3}\left( {{1 \over 2}CV_0^2} \right) = \frac{1}{6}CV_0^2$ This matches our calculated result.

Let's check the other options: B: $3(\frac{1}{2}CV_0^2) = \frac{3}{2}CV_0^2$ C: $3CV_0^2$ D: $\frac{1}{2}CV_0^2$

Our calculated value $E_D = \frac{1}{6}CV_0^2$ matches Option A.