A 500 W \(200 \mathrm{~V}\) resistive lamp is placed in series with an inductance to enable it to work from a 250 V \(50 \mathrm{~Hz}\) supply. What capacitor connected across the supply will make the power factor of the circuit unity?

AC circuit analysis involves understanding how alternating current (AC) flows through various electrical components and the effects these components have on the current's behavior. In the given exercise, we analyze a series circuit containing a resistor (the lamp) and an inductor. The goal is to adjust this circuit, which has a certain amount of inductive reactance, so that it can operate efficiently on an AC power source at a voltage different from the lamp's rating.

The process begins by calculating the current that flows through the lamp using the power formula, taking into account that the lamp operates on a 200V source. The impedance of the lamp is determined next, providing insight into its opposition to the AC. This step is critical, as impedance affects the current and voltage phase relationship—key factors in AC analysis. Once the individual components' reactances are known, we can understand how they affect the overall impedance of the circuit and, by extension, its power factor.

Electrical impedance represents the total opposition a circuit presents to the flow of alternating current. It is a complex value composed of resistance (the real part) and reactance (the imaginary part), which includes both inductive and capacitive components. In the context of our exercise, we first determined the impedance of the lamp by using Ohm's law. This step is pivotal in understanding how the lamp inherently resists the alternating current flow.

Impedance is crucial in AC circuits because it determines the phase difference between voltage and current. This phase difference is responsible for reactive power, which can lead to inefficiencies in power systems. Hence, accurately calculating impedance is essential for effective circuit analysis and subsequent steps in power factor correction.

Reactive power management concerns the balancing of reactive power generated by inductors and capacitors in an AC circuit. In the provided exercise, a purely resistive lamp and an inductor create an imbalance, manifesting as a phase difference between voltage and current, which results in a power factor less than one. This sub-optimal power factor reflects inefficiency where not all the power being consumed is being used for useful work.

To manage reactive power effectively and correct the power factor to unity (meaning voltage and current are in phase), we introduce a capacitor with an appropriate capacitive reactance. This reactance counteracts the effects of the inductive reactance, thereby reducing the reactive power and optimizing the circuit's efficiency. Managing reactive power is critical in electrical systems to minimize losses, reduce utility charges, and improve system stability.

Capacitive reactance is a measure of a capacitor's opposition to the change of voltage in an AC circuit. It depends inversely on both the frequency of the AC signal and the capacitance of the capacitor. In our exercise, we needed to calculate the capacitive reactance necessary to correct the power factor to unity. By introducing a capacitor with precise capacitive reactance, the inductive reactance brought by the inductor is neutralized, achieving the desired change in the circuit's dynamics.

The formula for capacitive reactance is given by \(X_C = \frac{1}{2\pi fC}\), where \(f\) is the frequency and \(C\) is the capacitance. By manipulating this relationship, we find the necessary capacitance to produce the needed reactance. The introduction of a capacitor with a calculated reactance ensures the efficient operation of the AC circuit, demonstrating the practical significance of capacitive reactance in power factor correction.