Question

A $100\, \Omega$ resistance and a capacitor of $100\, \Omega$ reactance are connected in series across a $220\, V$ source. When the capacitor is $50\%$ charged, the peak value of the displacement current is

• A. 2.2 A
• B. 11 A
• C. 4.4 A
• D. $11 \sqrt{2} A$

Answer 1

The Correct Option is A

\(X_{c} = 100\)
\(R = 100\)
\(Z = \sqrt{\left(100\right)^{2} + \left(100\right)^{2}} = 100 \sqrt{2}\) 
\(I = \frac{E}{Z} = \frac{220}{100\sqrt{2}} = \frac{2.2}{\sqrt{2}}\)

\(I_{0} = \sqrt{2} I\)
\(I_0 = \frac{2.2}{\sqrt{2}} \times \sqrt{2}\)
\(I_0 = 2.2\ \text{A}\)

So, the correct option is (A): 2.2 A

Answer 2

Given:
Resistance $R = 100\, \Omega$,
Capacitive reactance $X_C = 100\, \Omega$, 
Supply voltage $V = 220\, V$

Since $R = X_C$, the circuit is a series RC circuit with equal resistance and capacitive reactance.

Step 1: Calculate impedance
\(Z = \sqrt{R^2 + X_C^2} = \sqrt{100^2 + 100^2} = \sqrt{2} \times 100 \approx 141.4\, \Omega\)

Step 2: Calculate peak voltage
Given voltage is the rms voltage:
\(V_{\text{rms}} = 220\, V\)
So the peak voltage is:
\(V_0 = \sqrt{2} \cdot V_{\text{rms}} = \sqrt{2} \cdot 220 \approx 311\, V\)

Step 3: Displacement current in capacitor
In an AC circuit, the peak displacement current through a capacitor is:
\(I_0 = \frac{V_0}{X_C} = \frac{311}{100} = 3.11\, A\)

But the capacitor is 50% charged, so the voltage across it is half of peak:
\(V_C = \frac{V_0}{2} = \frac{311}{2} = 155.5\, V\)
So the instantaneous displacement current at this point is:
\(I = \frac{V_C}{X_C} = \frac{155.5}{100} = \boxed{1.555\, A}\)

But the question asks for the peak value of displacement current when the capacitor is 50% charged, which means we consider current when voltage is rising or falling.

Instead, in an AC circuit, the displacement current is maximum when voltage across capacitor is zero and minimum when capacitor is fully charged.
So when capacitor is 50% charged, the current is not maximum — but if this is interpreted for an AC sinusoidal steady state, then:

Actually, since it's a steady-state AC, and we are not integrating time, displacement current equals conduction current, and maximum (peak) current in the circuit is:
\(I_0 = \frac{V_0}{Z} = \frac{311}{141.4} \approx 2.2\, A\)

Final Answer:
\(\boxed{2.2\, A}\)