A parallel plate capacitor of capacitance 20 `mu F` 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 `mu A `
C. `60 mu A , 60 mu A `
D. `60 mu A , ` zero

Explanation

To solve the problem of finding the conduction current and displacement current in a parallel plate capacitor being charged by a voltage source, we can follow these steps:

### Step-by-Step Solution:

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

2. **Formula for Conduction Current**:
- The conduction current \( I_C \) through the capacitor can be calculated using the formula:
\[ I_C = C \frac{dV}{dt} \]

3. **Substituting Values**:
- Substitute the known values into the formula:
\[ I_C = (20 \times 10^{-6} \, F) \times (3 \, V/s) \]

4. **Calculate Conduction Current**:
- Perform the multiplication:
\[ I_C = 60 \times 10^{-6} \, A = 60 \, \mu A \]

5. **Displacement Current**:
- According to Maxwell's equations, the displacement current \( I_D \) is equal to the conduction current in a capacitor. Therefore:
\[ I_D = I_C = 60 \, \mu A \]

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

### Summary of Results:
- Conduction Current \( I_C = 60 \, \mu A \)
- Displacement Current \( I_D = 60 \, \mu A \)