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Step-by-Step Solution
Step 1: Identify the components and the given data
• A resistor of resistance R = 100 Ω and a capacitor with reactance XC = 100 Ω are connected in series.
• The RMS voltage across the series combination is 220 V.
• We want the peak (maximum) value of the displacement current when the capacitor is 50% charged.
Step 2: Calculate the impedance of the series circuit
For an RC series circuit, the net impedance Z is given by the expression:
$Z = \sqrt{R^2 + X_C^2}$
Substitute R = 100 Ω and XC = 100 Ω:
$Z = \sqrt{(100)^2 + (100)^2} = \sqrt{10000 + 10000} = 100\sqrt{2} \,\Omega.$
Step 3: Relate RMS voltage to peak voltage
For a sinusoidal source with RMS value Vrms, the peak voltage V0 is given by:
$V_{0} = \sqrt{2} \times V_{\mathrm{rms}}$
Here, Vrms = 220 V, so:
$V_{0} = \sqrt{2} \times 220 = 220\sqrt{2}\, \text{V}.$
Step 4: Find the peak current in the circuit
Ohm’s law for AC in terms of peak values gives:
$I_{0} = \frac{V_{0}}{Z}.$
Substitute V0 = 220√2 V and Z = 100√2 Ω:
$I_{0} = \frac{220\sqrt{2}}{100\sqrt{2}} = 2.2\, \text{A}.$
Step 5: Interpret the displacement current
In an AC circuit with a capacitor, when the capacitor is charging, the conduction current in the resistor and the displacement current in the capacitor are equal in magnitude. Therefore, the peak value of the displacement current is the same as the peak conduction current found in Step 4, which is 2.2 A.
Final Answer: 2.2 A
