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

 To solve this problem, we first need to understand the relationship between the given components in the circuit, namely the resistor and the capacitor, and how they interact with the AC voltage source.

The given problem involves a resistor with a resistance of \(100\, \Omega\) and a capacitor with a capacitive reactance of \(100\, \Omega\), connected in series across a voltage source of \(220\, V\). We are tasked with finding the peak displacement current when the capacitor is \(50\%\) charged.

1. The expression for the reactance of a capacitor is \(X_C = \frac{1}{\omega C}\), but since we've been directly given the capacitive reactance, we use this directly in our calculations.
2. In an AC circuit, the impedance \((Z)\) of components in series is given by: \(Z = \sqrt{R^2 + X_C^2}\), where \(R\) is the resistance and \(X_C\) is the capacitive reactance.
3. Substituting the given values, we have: \(Z = \sqrt{100^2 + 100^2} = \sqrt{20000} = 100\sqrt{2}\) ohms.
4. The peak voltage \((V_{\text{peak}})\) from the RMS voltage is: \(V_{\text{peak}} = V_{\text{rms}} \times \sqrt{2} = 220 \times \sqrt{2} \approx 311\, V\).
5. The peak current \((I_{\text{peak}})\) can be found using Ohm's Law for AC circuits: \(I_{\text{peak}} = \frac{V_{\text{peak}}}{Z} = \frac{311}{100\sqrt{2}} = \frac{311}{141.42} \approx 2.2\, A\).
6. This calculated peak current of \(2.2\, A\) represents the peak value of the displacement current when the capacitor is \(50\%\) charged.

Thus, the correct answer is 2.2 A.