91Ó°ÊÓ

Inductive reactance is a crucial concept in AC (Alternating Current) circuits that helps us understand how inductors behave when connected to an AC source. Inductors resist changes in current, and this resistance is uniquely characterized as reactance rather than a straightforward resistance. The formula for inductive reactance is given by:

• \(X_L = 2\pi f L\)

where:

• \(X_L\) is the inductive reactance measured in ohms (\(\Omega\)).
• \(f\) is the frequency of the AC source in hertz (Hz).
• \(L\) is the inductance value of the inductor given in henries (H).

In the original exercise, we have a unique combination of elements with a 35 mH inductance calculating its inductive reactance:

• \(X_L = 2 \times 3.14 \times 60 \times 35 \times 10^{-3}\)
• Thus, \(X_L \approx 13.2\, \Omega\).

This value of 13.2Ω acts almost like a 'resistance' but only to alternating currents.

Just as inductors have inductive reactance, capacitors exhibit something known as capacitive reactance. This is how much a capacitor resists an AC current based on its frequency and capacitance. Here's the formula you'll often use for this:

• \(X_C = \frac{1}{2\pi f C}\)

where:

• \(X_C\) stands for capacitive reactance also measured in ohms (\(\Omega\)).
• \(f\) represents the frequency of the AC supply in hertz (Hz).
• \(C\) is the capacitance of the capacitor in farads (F).

In solving the exercise, a 26 microfarad capacitor is used, and its capacitive reactance is calculated as:

• \(X_C = \frac{1}{2 \cdot 3.14 \cdot 60 \cdot 26 \times 10^{-6}}\)
• This results in \(X_C \approx 102.0\, \Omega\).

A larger reactance indicates the capacitor is good at blocking the current, while a smaller reactance means it's more permissive.

The phase angle in an AC circuit signifies the angular difference between current and voltage waveforms. In a purely resistive circuit, voltage and current are in phase, meaning they reach their maximum and minimum values simultaneously. However, with inductive or capacitive components in the circuit, this is not the case.
The phase angle \(\phi\) arises due to these reactive components, causing a time shift between the waveforms. You can calculate it using:

• \(\tan \phi = \frac{X_L - X_C}{R}\)

where:

• \(\phi\) is the phase angle in degrees (°).
• \(X_L\) is the inductive reactance.
• \(X_C\) is the capacitive reactance.
• \(R\) is the resistance in the circuit.

In the exercise, the calculation gives:

• \(\tan \phi = \frac{13.2 - 102.0}{2} = -44.4\)

Through the inverse tangent function, we find:

• \(\phi \approx -88.7^\circ\)

This indicates that the circuit’s current lags significantly behind the voltage, almost entirely out of phase, due to the dominance of capacitive reactance over inductive reactance.