A parallel plate capacitor 20 `muF` is being charged by a voltage source whose potential is changing at the rate of 3 V/s. The conduction current through the connecting wires, and the displacement current through the plates of the capacitor, would be, respectively:

Options

A. zero,zero
B. zero, `60 muA`
C. `60 muA, 60muA`
D. `60 muA`, zero

Explanation

To solve the problem, we need to find both the conduction current through the connecting wires and the displacement current through the plates of the capacitor.

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitance of the capacitor, \( C = 20 \, \mu F = 20 \times 10^{-6} \, F \)
- Rate of change of voltage, \( \frac{dV}{dt} = 3 \, V/s \)

2. **Use the formula for displacement current:**
The displacement current \( I_d \) through a capacitor can be calculated using the formula:
\[ I_d = C \frac{dV}{dt} \]

3. **Substituting the values into the formula:**
\[ I_d = 20 \times 10^{-6} \, F \times 3 \, V/s \]

4. **Calculate the displacement current:**
\[ I_d = 60 \times 10^{-6} \, A = 60 \, \mu A \]

5. **Conduction current:**
In a charging capacitor, the conduction current \( I_c \) is equal to the displacement current \( I_d \) when the capacitor is charging. Therefore:
\[ I_c = I_d = 60 \, \mu A \]

### Final Answer:
- The conduction current through the connecting wires is \( 60 \, \mu A \).
- The displacement current through the plates of the capacitor is also \( 60 \, \mu A \).