2312. Reactance, Inductive and CapacitiveLearning Objectives
Many circuits also contain capacitors and inductors, in addition to resistors and an AC voltage source. We have seen how capacitors and inductors respond to DC voltage when it is switched on and off. We will now explore how inductors and capacitors react to sinusoidal AC voltage. Inductors and Inductive ReactanceSuppose an inductor is connected directly to an AC voltage source, as shown in Figure 1. 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 Figure 1(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: AC Voltage in an InductorWhen a sinusoidal voltage is applied to an inductor, the voltage leads the current by onefourth of a cycle, or by a $\text{90\xba}$ phase angle. 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: $$I=\frac{V}{{X}_{L}}\text{,}$$where $V$ is the rms voltage across the inductor and ${X}_{L}$ is defined to be $${X}_{L}=\mathrm{2\pi}\text{fL}\text{,}$$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 Example 1: Calculating Inductive Reactance and then Current(a) Calculate the inductive reactance of a 3.00 mH inductor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What is the rms current at each frequency if the applied rms voltage is 120 V? Strategy The inductive reactance is found directly from the expression ${X}_{L}=\mathrm{2\pi}\text{fL}$. Once ${X}_{L}$ has been found at each frequency, Ohm’s law as stated in the Equation $I=V/{X}_{L}$ can be used to find the current at each frequency. Solution for (a) Entering the frequency and inductance into Equation ${X}_{L}=\mathrm{2\pi}\text{fL}$ gives $${X}_{L}=\mathrm{2\pi}\text{fL}=6.28(\text{60.0}/\text{s})(\mathrm{3.00\; mH})=\text{1.13 \Omega at60 Hz}.$$Similarly, at 10 kHz, $${X}_{L}=\mathrm{2\pi}\text{fL}=6\text{.}\text{28}(1.00\times {\text{10}}^{\text{4}}\text{/s})(3\text{.}\text{00 mH})=\text{188 \Omega at 10 kHz}.$$Solution for (b) The rms current is now found using the version of Ohm’s law in Equation $I=V/{X}_{L}$, given the applied rms voltage is 120 V. For the first frequency, this yields $$I=\frac{V}{{X}_{L}}=\frac{\text{120 V}}{\mathrm{1.13\; \Omega}}=\text{106 A at 60 Hz}.$$Similarly, at 10 kHz, $$I=\frac{V}{{X}_{L}}=\frac{\text{120 V}}{\text{188 \Omega}}=\mathrm{0.637\; A\; at\; 10\; kHz}.$$Discussion The inductor reacts very differently at the two different frequencies. At the higher frequency, its reactance is large and the current is small, consistent with how an inductor impedes rapid change. Thus high frequencies are impeded the most. Inductors can be used to filter out high frequencies; for example, a large inductor can be put in series with a sound reproduction system or in series with your home computer to reduce highfrequency sound output from your speakers or highfrequency power spikes into your computer. 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 ReactanceConsider the capacitor connected directly to an AC voltage source as shown in Figure 2. 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 Figure 2 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 onefourth of a cycle: AC Voltage in a CapacitorWhen a sinusoidal voltage is applied to a capacitor, the voltage follows the current by onefourth of a cycle, or by a $\text{90\xba}$ phase angle. 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 $$I=\frac{V}{{X}_{C}}\text{,}$$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 $${X}_{C}=\frac{1}{\mathrm{2\pi}\text{fC}}\text{,}$$where ${X}_{C}$ is called the Example 2: Calculating Capacitive Reactance and then Current(a) Calculate the capacitive reactance of a 5.00 mF capacitor when 60.0 Hz and 10.0 kHz AC voltages are applied. (b) What is the rms current if the applied rms voltage is 120 V? Strategy The capacitive reactance is found directly from the expression in ${X}_{C}=\frac{1}{\mathrm{2\pi}\text{fC}}$. Once ${X}_{C}$ has been found at each frequency, Ohm’s law stated as $I=V/{X}_{C}$ can be used to find the current at each frequency. Solution for (a) Entering the frequency and capacitance into ${X}_{C}=\frac{1}{\mathrm{2\pi}\text{fC}}$ gives $$\begin{array}{lll}{X}_{C}& =& \frac{1}{\mathrm{2\pi}\text{fC}}\\ & =& \frac{1}{6\text{.}\text{28}(\text{60}\text{.}0/\text{s})(5\text{.}\text{00}\mu \text{F})}=\text{531 \Omega at60 Hz}\text{.}\end{array}$$Similarly, at 10 kHz, $$\begin{array}{lll}{X}_{C}& =& \frac{1}{\mathrm{2\pi}\text{fC}}=\frac{1}{6\text{.}\text{28}(1\text{.}\text{00}\times {\text{10}}^{4}/\text{s})(5\text{.}\text{00}\mu \text{F})}\\ & =& \text{3.18 \Omega at10 kHz}\end{array}\text{.}$$Solution for (b) The rms current is now found using the version of Ohm’s law in $I=V/{X}_{C}$, given the applied rms voltage is 120 V. For the first frequency, this yields $$I=\frac{V}{{X}_{C}}=\frac{\text{120 V}}{\text{531 \Omega}}=\text{0.226 A at 60 Hz}.$$Similarly, at 10 kHz, $$I=\frac{V}{{X}_{C}}=\frac{\text{120 V}}{\text{3.18 \Omega}}=\mathrm{37.7\; A\; at\; 10\; kHz}.$$Discussion The capacitor reacts very differently at the two different frequencies, and in exactly the opposite way an inductor reacts. At the higher frequency, its reactance is small and the current is large. Capacitors favor change, whereas inductors oppose change. Capacitors impede low frequencies the most, since low frequency allows them time to become charged and stop the current. Capacitors can be used to filter out low frequencies. For example, a capacitor in series with a sound reproduction system rids it of the 60 Hz hum. 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 CircuitJust as a reminder, consider Figure 3, 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: AC Voltage in a ResistorWhen a sinusoidal voltage is applied to a resistor, the voltage is exactly in phase with the current—they have a $\text{0\xba}$ phase angle. Section Summary
Conceptual QuestionsExercise 1Presbycusis is a hearing loss due to age that progressively affects higher frequencies. A hearing aid amplifier is designed to amplify all frequencies equally. To adjust its output for presbycusis, would you put a capacitor in series or parallel with the hearing aid’s speaker? Explain. Exercise 2Would you use a large inductance or a large capacitance in series with a system to filter out low frequencies, such as the 100 Hz hum in a sound system? Explain. Exercise 3Highfrequency noise in AC power can damage computers. Does the plugin unit designed to prevent this damage use a large inductance or a large capacitance (in series with the computer) to filter out such high frequencies? Explain. Exercise 4Does inductance depend on current, frequency, or both? What about inductive reactance? Exercise 5Explain why the capacitor in Figure 4(a) acts as a lowfrequency filter between the two circuits, whereas that in Figure 4(b) acts as a highfrequency filter. Exercise 6If the capacitors in Figure 4 are replaced by inductors, which acts as a lowfrequency filter and which as a highfrequency filter? Problems & ExercisesExercise 1At what frequency will a 30.0 mH inductor have a reactance of $\text{100 \Omega}$? Show/Hide Solution Solution531 Hz Exercise 2What value of inductance should be used if a $\text{20.0 k\Omega}$ reactance is needed at a frequency of 500 Hz? Exercise 3What capacitance should be used to produce a $\text{2.00 M\Omega}$ reactance at 60.0 Hz? Show/Hide Solution Solution1.33 nF Exercise 4At what frequency will an 80.0 mF capacitor have a reactance of $\text{0.250 \Omega}$? Exercise 5(a) Find the current through a 0.500 H inductor connected to a 60.0 Hz, 480 V AC source. (b) What would the current be at 100 kHz? Show/Hide Solution Solution(a) 2.55 A (b) 1.53 mA Exercise 6(a) What current flows when a 60.0 Hz, 480 V AC source is connected to a $\text{0.250 \mu F}$ capacitor? (b) What would the current be at 25.0 kHz? Exercise 7A 20.0 kHz, 16.0 V source connected to an inductor produces a 2.00 A current. What is the inductance? Show/Hide Solution Solution$\text{63.7 \xb5H}$ Exercise 8A 20.0 Hz, 16.0 V source produces a 2.00 mA current when connected to a capacitor. What is the capacitance? Exercise 9(a) An inductor designed to filter highfrequency noise from power supplied to a personal computer is placed in series with the computer. What minimum inductance should it have to produce a $\text{2.00 k\Omega}$ reactance for 15.0 kHz noise? (b) What is its reactance at 60.0 Hz? Show/Hide Solution Solution(a) 21.2 mH (b) $\text{8.00 \Omega}$ Exercise 10The capacitor in Figure 4(a) is designed to filter lowfrequency signals, impeding their transmission between circuits. (a) What capacitance is needed to produce a $\text{100 k\Omega}$ reactance at a frequency of 120 Hz? (b) What would its reactance be at 1.00 MHz? (c) Discuss the implications of your answers to (a) and (b). Exercise 11The capacitor in Figure 4(b) will filter highfrequency signals by shorting them to earth/ground. (a) What capacitance is needed to produce a reactance of $\text{10.0 m\Omega}$ for a 5.00 kHz signal? (b) What would its reactance be at 3.00 Hz? (c) Discuss the implications of your answers to (a) and (b). Show/Hide Solution Solution(a) 3.18 mF (b) $\text{16.7 \Omega}$ Exercise 12Unreasonable Results In a recording of voltages due to brain activity (an EEG), a 10.0 mV signal with a 0.500 Hz frequency is applied to a capacitor, producing a current of 100 mA. Resistance is negligible. (a) What is the capacitance? (b) What is unreasonable about this result? (c) Which assumption or premise is responsible? Exercise 13Construct Your Own Problem Consider the use of an inductor in series with a computer operating on 60 Hz electricity. Construct a problem in which you calculate the relative reduction in voltage of incoming high frequency noise compared to 60 Hz voltage. Among the things to consider are the acceptable series reactance of the inductor for 60 Hz power and the likely frequencies of noise coming through the power lines.
