1. Given data

• Capacitance: $C = 200\,\text{pF} = 200 \times 10^{-12}\,\text{F} = 2 \times 10^{-10}\,\text{F}$
• RMS voltage: $V_{\text{rms}} = 230\,\text{V}$
• Angular frequency: $\omega = 300\,\text{rad s}^{-1}$

We need the rms conduction current in the circuit and displacement current in the capacitor.

2. Current in a capacitor in AC

For a capacitor connected to AC,

$I_{\text{rms}} = V_{\text{rms}} \cdot \omega C$

because capacitive reactance is

$X_C = \frac{1}{\omega C}$

and hence

$I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = V_{\text{rms}}\,\omega C$

So,

$I_{\text{rms}} = 230 \times 300 \times 2 \times 10^{-10}$

$= 230 \times 6 \times 10^{-8}$

$= 1380 \times 10^{-8}$

$= 1.38 \times 10^{-5}\,\text{A}$

$= 13.8 \times 10^{-6}\,\text{A}$

$= 13.8\,\mu\text{A}$

3. Conduction current and displacement current

In a capacitor circuit:

• The current in the connecting wires is the conduction current.
• Between the plates, the equivalent current is the displacement current.

By Maxwell’s theory, in a capacitor these two are equal at every instant, and therefore their rms values are also equal.

Hence,

$I_{\text{conduction,rms}} = I_{\text{displacement,rms}} = 13.8\,\mu\text{A}$

4. Checking options

• A: $14.3\,\mu\text{A}$ and $143\,\mu\text{A}$ ❌
• B: $13.8\,\mu\text{A}$ and $13.8\,\mu\text{A}$ ✅
• C: $13.8\,\mu\text{A}$ and $138\,\mu\text{A}$ ❌
• D: $1.38\,\mu\text{A}$ and $1.38\,\mu\text{A}$ ❌

So the correct option is B.