Question

$100\Omega$ resistance and a capacitor of $100\Omega$ reactance are connected in series across a $220\text{V}$ source. The peak value of the displacement current is

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

Answer 1

The correct option is A $2.2\text{A}$

Given:-
$R=100\Omega ;X_C=100\Omega ;V_{rms}=220\text{V}$

The net impedance of the circuit is

$Z=\sqrt{R^2+X{{}_C}^2}$

$Z=\sqrt{{100}^2+{100}^2}$

$Z=100\sqrt{2}\Omega$

We know that the peak value of displacement current is equal to the Peak value of conduction current

$\therefore (I_d{)}_{peak}=(I_C{)}_{peak}=(I_C{)}_{rms}\sqrt{2}$

$(I_d{)}_{peak}=(I_C{)}_{peak}=\frac{V_{rms}}{Z}\sqrt{2}$

$(I_d{)}_{peak}=\frac{220\times \sqrt{2}}{100\sqrt{2}}$

$\therefore (I_d{)}_{peak}=2.2\text{A}$

Hence, option $\left(\text{A}\right)$ is correct.