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

To illustrate this relationship graphically:
In an AC circuit, a capacitor opposes current flow based on the AC supply frequency. Higher frequencies lead to lower reactance, permitting more current to pass, a property known as capacitive reactance.
In a DC circuit, a capacitor initially conducts during charging. Once fully charged, it acts as an open circuit, halting current flow in a steady-state DC condition.
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.
AB is a part of an electrical circuit (see figure). The potential difference \(V_A - V_B\), at the instant when current \(i = 2\) A and is increasing at a rate of 1 amp/second is: