The circuit element ‘X’ where the current leads the voltage by \( \frac{\pi}{2} \) is the capacitor. In capacitive circuits, the charging current for the capacitor leads the voltage across the capacitor.
The reactance \( X_C \) of a capacitor is given by:
\[ X_C = \frac{1}{\omega C} \]
where \( \omega \) is the angular frequency (\( \omega = 2 \pi f \), with \( f \) being the frequency of the AC supply), and \( C \) is the capacitance in farads.
The capacitive reactance \( X_C \), for a capacitor in an AC circuit is defined by the formula:
\[ X_C = \frac{1}{\omega C} \]
where \( \omega = 2 \pi f \) represents the angular frequency, and \( C \) is the capacitance. This formula shows that \( X_C \) is inversely proportional to the frequency \( f \).

To illustrate the relationship graphically:
In an AC circuit, a capacitor impedes the flow of the current depending on the frequency of the AC supply. Higher frequencies reduce the reactance, allowing more current to pass through, demonstrating a characteristic called capacitive reactance.
In a DC circuit, a capacitor initially conducts as it charges, but once fully charged, it acts as an open circuit. Thus, after the initial charge period, no current flows through the capacitor in a steady-state DC circuit.
A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2 \(\Omega\) then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is _____ N.
