Inductors and Inductive Reactance

Suppose an inductor is connected directly to an AC voltage source, as shown in [link]. It is reasonable to assume negligible resistance, since in practice we can make the resistance of an inductor so small that it has a negligible effect on the circuit. Also shown is a graph of voltage and current as functions of time.

The graph in [link](b) starts with voltage at a maximum. Note that the current starts at zero and rises to its peak after the voltage that drives it, just as was the case when DC voltage was switched on in the preceding section. When the voltage becomes negative at point a, the current begins to decrease; it becomes zero at point b, where voltage is its most negative. The current then becomes negative, again following the voltage. The voltage becomes positive at point c and begins to make the current less negative. At point d, the current goes through zero just as the voltage reaches its positive peak to start another cycle. This behavior is summarized as follows:

Current lags behind voltage, since inductors oppose change in current. Changing current induces a back emf $V=-L(\mathrm{\Delta }I/\mathrm{\Delta }t)$. This is considered to be an effective resistance of the inductor to AC. The rms current $I$ through an inductor $L$ is given by a version of Ohm’s law:

where $V$ is the rms voltage across the inductor and $X_L$ is defined to be

with $f$ the frequency of the AC voltage source in hertz (An analysis of the circuit using Kirchhoff’s loop rule and calculus actually produces this expression). $X_L$ is called the inductive reactance, because the inductor reacts to impede the current. $X_L$ has units of ohms ($\mathrm{1\; H}=\mathrm{1\; \Omega }\cdot \text{s}$, so that frequency times inductance has units of $(\text{cycles/s})(\Omega \cdot \text{s})=\Omega$), consistent with its role as an effective resistance. It makes sense that $X_L$ is proportional to $L$, since the greater the induction the greater its resistance to change. It is also reasonable that $X_L$ is proportional to frequency $f$, since greater frequency means greater change in current. That is, $\mathrm{\Delta }I\mathrm{/\Delta }t$ is large for large frequencies (large $f$, small $\mathrm{\Delta }t$). The greater the change, the greater the opposition of an inductor.

Note that although the resistance in the circuit considered is negligible, the AC current is not extremely large because inductive reactance impedes its flow. With AC, there is no time for the current to become extremely large.

Capacitors and Capacitive Reactance

Consider the capacitor connected directly to an AC voltage source as shown in [link]. The resistance of a circuit like this can be made so small that it has a negligible effect compared with the capacitor, and so we can assume negligible resistance. Voltage across the capacitor and current are graphed as functions of time in the figure.

The graph in [link] starts with voltage across the capacitor at a maximum. The current is zero at this point, because the capacitor is fully charged and halts the flow. Then voltage drops and the current becomes negative as the capacitor discharges. At point a, the capacitor has fully discharged ($Q=0$ on it) and the voltage across it is zero. The current remains negative between points a and b, causing the voltage on the capacitor to reverse. This is complete at point b, where the current is zero and the voltage has its most negative value. The current becomes positive after point b, neutralizing the charge on the capacitor and bringing the voltage to zero at point c, which allows the current to reach its maximum. Between points c and d, the current drops to zero as the voltage rises to its peak, and the process starts to repeat. Throughout the cycle, the voltage follows what the current is doing by one-fourth of a cycle:

The capacitor is affecting the current, having the ability to stop it altogether when fully charged. Since an AC voltage is applied, there is an rms current, but it is limited by the capacitor. This is considered to be an effective resistance of the capacitor to AC, and so the rms current $I$ in the circuit containing only a capacitor $C$ is given by another version of Ohm’s law to be

where $V$ is the rms voltage and $X_C$ is defined (As with $X_L$, this expression for $X_C$ results from an analysis of the circuit using Kirchhoff’s rules and calculus) ^{to be}

where $X_C$ is called the capacitive reactance, because the capacitor reacts to impede the current. $X_C$ has units of ohms (verification left as an exercise for the reader). $X_C$ is inversely proportional to the capacitance $C$; the larger the capacitor, the greater the charge it can store and the greater the current that can flow. It is also inversely proportional to the frequency $f$; the greater the frequency, the less time there is to fully charge the capacitor, and so it impedes current less.

Although a capacitor is basically an open circuit, there is an rms current in a circuit with an AC voltage applied to a capacitor. This is because the voltage is continually reversing, charging and discharging the capacitor. If the frequency goes to zero (DC), $X_C$ tends to infinity, and the current is zero once the capacitor is charged. At very high frequencies, the capacitor’s reactance tends to zero—it has a negligible reactance and does not impede the current (it acts like a simple wire). Capacitors have the opposite effect on AC circuits that inductors have.

Resistors in an AC Circuit

Just as a reminder, consider [link], which shows an AC voltage applied to a resistor and a graph of voltage and current versus time. The voltage and current are exactly in phase in a resistor. There is no frequency dependence to the behavior of plain resistance in a circuit: