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

Options

A. 2.2 A
B. 0.45833333333333
C. 4.4 A
D. `11sqrt(2)` A

Explanation

To solve the problem step by step, we need to find the peak value of the displacement current when a 100 Ω resistance and a capacitor with 100 Ω reactance are connected in series across a 220 V source.

### Step 1: Calculate the peak voltage (V_max)
The given voltage is the RMS voltage (V_rms = 220 V). To find the peak voltage (V_max), we use the formula:

\[ V_{max} = V_{rms} \times \sqrt{2} \]

Substituting the value:

\[ V_{max} = 220 \times \sqrt{2} \approx 220 \times 1.414 \approx 311.13 \, V \]

### Step 2: Calculate the impedance (Z)
In a series circuit with resistance (R) and capacitive reactance (X_C), the total impedance (Z) is given by:

\[ Z = \sqrt{R^2 + X_C^2} \]

Here, R = 100 Ω and X_C = 100 Ω. Thus,

\[ Z = \sqrt{100^2 + 100^2} = \sqrt{10000 + 10000} = \sqrt{20000} = 100\sqrt{2} \, \Omega \]

### Step 3: Calculate the peak current (I_max)
The peak current (I_max) can be calculated using Ohm's law for AC circuits:

\[ I_{max} = \frac{V_{max}}{Z} \]

Substituting the values we found:

\[ I_{max} = \frac{311.13}{100\sqrt{2}} \approx \frac{311.13}{141.42} \approx 2.20 \, A \]

### Step 4: Conclusion
The peak value of the displacement current when the capacitor is 50% charged is approximately **2.20 A**.