32.3 $R C$ Filter Circuits

You learned in Chapter 28 that a resistance $R$ causes a capacitor to be charged or discharged with time constant $τ = R C$ . We called this an $R C$ circuit. Now that we’ve looked at resistors and capacitors individually, let’s explore what happens if an $R C$ circuit is driven continuously by an alternating current source.

FIGURE 32.10 shows a circuit in which a resistor $R$ and capacitor $C$ are in series with an emf $ε$ oscillating at angular frequency $ω$ . Before launching into a formal analysis, let’s try to understand qualitatively how this circuit will respond as the frequency is varied. If the frequency is very low, the capacitive reactance will be very large, and thus the peak current $I_{C}$ will be very small. The peak current through the resistor is the same as the peak current to and from the capacitor (just as in DC circuits, conservation of charge requires $I_{R} = I_{C}$ ); hence we expect the resistor’s peak voltage $V_{R} = I_{R} R$ to be very small at very low frequencies.

The figure illustrates a circuit diagram of a resistor and a capacitor in series with an "A" C voltage source.

FIGURE 32.10 Full Alternative Text

On the other hand, suppose the frequency is very high. Then the capacitive reactance approaches zero and the peak current, determined by the resistance alone, will be $I_{R} = ε_{0} / R$ . The resistor’s peak voltage $V_{R} = I R$ will approach the peak source voltage $ε_{0}$ at very high frequencies.

This reasoning leads us to expect that $V_{R}$ will increase steadily from 0 to $ε_{0}$ as $ω$ is increased from 0 to very high frequencies. Kirchhoff’s loop law has to be obeyed, so the capacitor voltage $V_{C}$ will decrease from $ε_{0}$ to 0 during the same change of frequency. A quantitative analysis will show us how this behavior can be used as a filter.

The goal of a quantitative analysis is to determine the peak current $I$ and the two peak voltages $V_{R}$ and $V_{C}$ as functions of the emf amplitude $ε_{0}$ and frequency $ω$ . Our analytic procedure is based on the fact that the instantaneous current $i$ is the same for two circuit elements in series.

Using phasors to analyze an $R C$ circuit

Full Alternative Text

Begin by drawing a current phasor of length $I$ . This is the starting point because the series circuit elements have the same current $i$ . The angle at which the phasor is drawn is not relevant.

Full Alternative Text

The current and voltage of a resistor are in phase, so draw a resistor voltage phasor of length $V_{R}$ parallel to the current phasor $I$ . The capacitor current leads the capacitor voltage by $90 °$ , so draw a capacitor voltage phasor of length $V_{C}$ that is $90 °$ behind [i.e., clockwise (cw) from] the current phasor.

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The series resistor and capacitor are in parallel with the emf, so their instantaneous voltages satisfy $v_{R} + v_{C} = ε$ . This is a vector addition of phasors, so draw the emf phasor as the vector sum of the two voltage phasors. The emf is $ε = ε_{0} cos ω t$ , hence the emf phasor is at angle $ω t$ .

Full Alternative Text

The length of the emf phasor, $ε_{0}$ , is the hypotenuse of a right triangle formed by the resistor and capacitor phasors. Thus $ε_{0}^{2} = V_{R}^{2} + V_{C}^{2}$ .

The relationship $ε_{0} ^{2} = V_{R} ^{2} + V_{C} ^{2}$ is based on the peak values, not the instantaneous values, because the peak values are the lengths of the sides of the right triangle. The peak voltages are related to the peak current $I$ via $V_{R} = I R$ and $V_{C} = I X_{C}$ , thus

\begin{array}{l} ε_{0}^{2} & = & V_{R}^{2} + V_{C}^{2} = {(I R)}^{2} + {(I X_{C})}^{2} = (R^{2} + X_{C}^{2}) I^{2} \\ = & (R^{2} + 1 / ω^{2} C^{2}) I^{2} \end{array}

(32.12)

Consequently, the peak current in the $R C$ circuit is

I = \frac{ε_{0}}{\sqrt{R^{2} + X_{C} ​^{2}}} = \frac{ε_{0}}{\sqrt{R^{2} + 1 / ω^{2} C^{2}}}

(32.13)

Knowing $I$ gives us the two peak voltages:

\begin{array}{l} V_{R} = I R = \frac{ε_{0} R}{\sqrt{R^{2} + X_{C} ​^{2}}} = \frac{ε_{0} R}{\sqrt{R^{2} + 1 / ω^{2} C^{2}}} \\ V_{C} = I X_{C} = \frac{ε_{0} X_{C}}{\sqrt{R^{2} + X_{C} ​^{2}}} = \frac{ε_{0} / ω C}{\sqrt{R^{2} + 1 / ω^{2} C^{2}}} \end{array}

(32.14)