15.3 RLC Series Circuits with AC

The ac circuit shown in Figure 15.11, called an RLC series circuit, is a series combination of a resistor, capacitor, and inductor connected across an ac source. It produces an emf of

$v open parenthesis t close parenthesis equals V sub 0 sin ω t.$

Figure a shows a circuit with an AC voltage source connected to a resistor, a capacitor and an inductor in series. The source is labeled V0 sine omega t. Figure b shows sine waves of AC voltage and current on the same graph. Voltage has a greater amplitude than current and its maximum value is marked V0 on the y axis. The maximum value of current is marked I0. The two curves have the same wavelength but are out of phase. The voltage curve is labeled V parentheses t parentheses equal to V0 sine omega t. The current curve is labeled I parentheses t parentheses equal to I0 sine parentheses omega t minus phi parentheses. — **Figure 15.11** (a) An *RLC* series circuit. (b) A comparison of the generator output voltage and the current. The value of the phase difference $ϕ$ depends on the values of R, C, and L.

Since the elements are in series, the same current flows through each element at all points in time. The relative phase between the current and the emf is not obvious when all three elements are present. Consequently, we represent the current by the general expression

$i (t) = I_{0} sin (ω t - ϕ),$

where $I sub 0$ is the current amplitude and $ϕ$ is the phase angle between the current and the applied voltage. The phase angle is thus the amount by which the voltage and current are out of phase with each other in a circuit. Our task is to find $I sub 0 and ϕ.$

A phasor diagram involving $i (t), v_{R} (t), v_{C} (t), and v_{L} (t)$ is helpful for analyzing the circuit. As shown in Figure 15.12, the phasor representing $v_{R} (t)$ points in the same direction as the phasor for $i (t);$ its amplitude is $V sub R equals I sub 0 R.$ The $v_{C} (t)$ phasor lags the i(t) phasor by $π / 2$ rad and has the amplitude $V sub C equals I sub 0 X sub C.$ The phasor for $v_{L} (t)$ leads the i(t) phasor by $π / 2$ rad and has the amplitude $V sub L equals I sub 0 X sub L.$

Figure shows the coordinate axes, with four arrows starting from the origin. Arrow V subscripts R points up and right, making an angle omega t minus phi with the x axis. Its y intercept is V subscript R parentheses t parentheses. Arrow I0 is along arrow V subscript R, but shorter than it. Arrow V subscript L points up and left and is perpendicular to V subscript R. It makes a y intercept V subscript L parentheses t parentheses. Arrow V subscript C points down and right. It is perpendicular to V subscript R. It makes a y intercept V subscript C parentheses t parentheses. Three arrows labeled omega are each perpendicular to V subscript R, V subscript L and V subscript C, shown near their tips. — **Figure 15.12** The phasor diagram for the *RLC* series circuit of Figure 15.11.

At any instant, the voltage across the RLC combination is $v_{R} (t) + v_{L} (t) + v_{C} (t) = v (t),$ the emf of the source. Since a component of a sum of vectors is the sum of the components of the individual vectors—for example, ${(A + B)}_{y} = A_{y} + B_{y}$ —the projection of the vector sum of phasors onto the vertical axis is the sum of the vertical projections of the individual phasors. Hence, if we add vectorially the phasors representing $v_{R} (t), v_{L} (t), and v_{C} (t)$ and then find the projection of the resultant onto the vertical axis, we obtain

$v_{R} (t) + v_{L} (t) + v_{C} (t) = v (t) = V_{0} sin ω t .$

The vector sum of the phasors is shown in Figure 15.13. The resultant phasor has an amplitude $V sub 0$ and is directed at an angle $ϕ$ with respect to the $v_{R} (t),$ or $i (t)$ , phasor. The projection of this resultant phasor onto the vertical axis is $v open parenthesis t close parenthesis equals V sub 0 sin ω t.$ We can easily determine the unknown quantities $I sub 0$ and $ϕ$ from the geometry of the phasor diagram. For the phase angle,

Four straight segments radiate from a common origin on axes marked in millivolts; they carry no arrowheads, so their direction is read from which way they point. One lies along the horizontal axis — the resistor voltage, which is also the direction of the current; one points straight up and one straight down, the inductor and capacitor voltages, exactly opposite one another because they sit a quarter cycle either side of the current. A heavier segment, ringed at its far end, runs diagonally to the corner of the faint rectangle those make — the source voltage, leaning above the horizontal because at the starting frequency the inductor wins. Lowering the frequency shortens the upward segment and stretches the downward one until, at resonance, the two cancel and the source segment lies flat along the axis at its full length; below resonance it swings under the axis and the current leads instead. Its tip rides a dashed arc the whole way, because the source amplitude never changes — only its angle does. Adjustable parameter: Drive frequency f (the worked example uses 200 Hz) (f) = 200 Hz. Viewing window: x from -97.29 to 197.29, y from -81.09 to 101.09. — The phasor construction this section builds, for the circuit the example An RLC Series Circuit uses: R = 4.00 Ω, L = 3.00 mH, C = 8.00 × 10⁻⁴ F, driven at amplitude V₀ = 0.100 V. Both axes are voltage in millivolts — the same unit on each, which is what makes the drawn angle the true phase angle. The segments are phasor AMPLITUDES drawn in the frame co-rotating with the current, that is, the snapshot in which the current phasor lies along the horizontal axis; they are not instantaneous voltages, and this section's own figures show the same phasors turned to a general angle ωt with the instantaneous value read off the vertical projection. Fixing the frame to the current removes ωt and leaves every angle between the phasors unchanged, which is what lets the frequency become a slider. At the example's 200 Hz the inductor wins — X_L = 3.77 Ω against X_C = 0.995 Ω — so V_L − V_C points up and the source leads the current by φ = 34.8°, the 0.607 rad the example computes. Drag the frequency down through 102.7 Hz and the diagram flips character: the reactances cancel at resonance, f₀ = 102.7 Hz, where V_R alone carries the whole 100 mV — the 1 Hz slider stops at 103 Hz, close enough that the residual tilt is a fraction of a pixel, and below it the capacitor dominates and the current leads instead. The dashed arc is the reason the picture stays readable: the source amplitude is set by the generator, so the resultant is always 100 mV long and only its angle can change.

$ϕ = {tan}^{−1} \frac{V_{L} - V_{C}}{V_{R}} = {tan}^{−1} \frac{I_{0} X_{L} - I_{0} X_{C}}{I_{0} R},$

and after cancellation of $I sub 0,$ this becomes

$ϕ = {tan}^{−1} \frac{X_{L} - X_{C}}{R} .$

(15.9)

Furthermore, from the Pythagorean theorem,

$V_{0} = \sqrt{V_{R}^{2} + {(V_{L} - V_{C})}^{2}} = \sqrt{{(I_{0} R)}^{2} + {(I_{0} X_{L} - I_{0} X_{C})}^{2}} = I_{0} \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} .$

Three arrows start from the origin on the coordinate axis. Arrow V subscript R points up and right, making an angle omega t minus phi with the x axis. Arrow V0 points up and right, making an angle omega t with the x axis. It makes an angle phi with the arrow V subscript R. It makes a y intercept labeled V0 sine omega t. The third arrow is labeled V subscript L minus V subscript C. It points up and left and is perpendicular to arrow V subscript R. Dotted lines indicate that the rectangle formed with its longer side being V subscript R and shorter side being V subscript L minus V subscript C, would have the arrow V0 as a diagonal. An arrow labeled omega is shown near the tip of V subscript R, perpendicular to it. — **Figure 15.13** The resultant of the phasors for $v_{L} (t)$ , $v_{C} (t)$ , and $v_{R} (t)$ is equal to the phasor for $v open parenthesis t close parenthesis equals V sub 0 sin ω t.$ The i(t) phasor (not shown) is aligned with the $v_{R} (t)$ phasor.

The current amplitude is therefore the ac version of Ohm’s law:

$I_{0} = \frac{V_{0}}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}} = \frac{V_{0}}{Z},$

(15.10)

where

$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$

(15.11)

is known as the impedance of the circuit. Its unit is the ohm, and it is the ac analog to resistance in a dc circuit, which measures the combined effect of resistance, capacitive reactance, and inductive reactance (Figure 15.14).

Photograph of power capacitors at a power station. — **Figure 15.14** Power capacitors are used to balance the impedance of the effective inductance in transmission lines.

The RLC circuit is analogous to the wheel of a car driven over a corrugated road (Figure 15.15). The regularly spaced bumps in the road drive the wheel up and down; in the same way, a voltage source increases and decreases. The shock absorber acts like the resistance of the RLC circuit, damping and limiting the amplitude of the oscillation. Energy within the wheel system goes back and forth between kinetic and potential energy stored in the car spring, analogous to the shift between a maximum current, with energy stored in an inductor, and no current, with energy stored in the electric field of a capacitor. The amplitude of the wheel’s motion is at a maximum if the bumps in the road are hit at the resonant frequency, which we describe in more detail in Resonance in an AC Circuit.

Figure shows one wheel of a car. Arrows show the up-down motion of its shock absorber spring. — **Figure 15.15** On a car, the shock absorber damps motion and dissipates energy. This is much like the resistance in an *RLC* circuit. The mass and spring determine the resonant frequency.

Problem-Solving Strategy

AC Circuits

To analyze an ac circuit containing resistors, capacitors, and inductors, it is helpful to think of each device’s reactance and find the equivalent reactance using the rules we used for equivalent resistance in the past. Phasors are a great method to determine whether the emf of the circuit has positive or negative phase (namely, leads or lags other values). A mnemonic device of “ELI the ICE man” is sometimes used to remember that the emf (E) leads the current (I) in an inductor (L) and the current (I) leads the emf (E) in a capacitor (C).

Use the following steps to determine the emf of the circuit by phasors:

Draw the phasors for voltage across each device: resistor, capacitor, and inductor, including the phase angle in the circuit.
If there is both a capacitor and an inductor, find the net voltage from these two phasors, since they are antiparallel.
Find the equivalent phasor from the phasor in step 2 and the resistor’s phasor using trigonometry or components of the phasors. The equivalent phasor found is the emf of the circuit.

Example 1 An RLC Series Circuit

The output of an ac generator connected to an RLC series combination has a frequency of 200 Hz and an amplitude of 0.100 V. If $R equals 4.00 Ω,$ $L = 3.00 \times 10^{−3} H,$ and $C = 8.00 \times 10^{−4} F,$ what are (a) the capacitive reactance, (b) the inductive reactance, (c) the impedance, (d) the current amplitude, and (e) the phase difference between the current and the emf of the generator?

Strategy The reactances and impedance in (a)–(c) are found by substitutions into Equation 15.3, Equation 15.8, and Equation 15.11, respectively. The current amplitude is calculated from the peak voltage and the impedance. The phase difference between the current and the emf is calculated by the inverse tangent of the difference between the reactances divided by the resistance.

Solution

From Equation 15.3, the capacitive reactance is
$X_{C} = \frac{1}{ω C} = \frac{1}{2 π (200 Hz) (8.00 \times 10^{−4} F)} = 0.995 Ω .$
From Equation 15.8, the inductive reactance is
$X_{L} = ω L = 2 π (200 Hz) (3.00 \times 10^{−3} H) = 3.77 Ω .$
Substituting the values of R, $X_{C}$ , and $X_{L}$ into Equation 15.11, we obtain for the impedance
$Z = \sqrt{{(4.00 Ω)}^{2} + {(3.77 Ω - 0.995 Ω)}^{2}} = 4.87 Ω .$
The current amplitude is
$I_{0} = \frac{V_{0}}{Z} = \frac{0.100 V}{4.87 Ω} = 2.05 \times 10^{−2} A .$
From Equation 15.9, the phase difference between the current and the emf is
$ϕ = {tan}^{−1} \frac{X_{L} - X_{C}}{R} = {tan}^{−1} \frac{2.77 Ω}{4.00 Ω} = 0.607 rad .$

Significance The phase angle is positive because the reactance of the inductor is larger than the reactance of the capacitor.

Summary

An RLC series circuit is a resistor, capacitor, and inductor series combination across an ac source.
The same current flows through each element of an RLC series circuit at all points in time.
The counterpart of resistance in a dc circuit is impedance, which measures the combined effect of resistors, capacitors, and inductors. The maximum current is defined by the ac version of Ohm’s law.
Impedance has units of ohms and is found using the resistance, the capacitive reactance, and the inductive reactance.

Conceptual Questions

In an RLC series circuit, can the voltage measured across the capacitor be greater than the voltage of the source? Answer the same question for the voltage across the inductor.

yes for both

Problems

What is the impedance of a series combination of a $50 -Ω$ resistor, a $5.0 - μ F$ capacitor, and a $10 - μ F$ capacitor at a frequency of 2.0 kHz?

A resistor and capacitor are connected in series across an ac generator. The emf of the generator is given by $v open parenthesis t close parenthesis equals V sub 0 cos ω t,$ where $V sub 0 equals 120 V,$ $ω equals 120 π rad/s,$ $R equals 400 Ω$ , and $C equals 4.0 μ F$ . (a) What is the impedance of the circuit? (b) What is the amplitude of the current through the resistor? (c) Write an expression for the current through the resistor. (d) Write expressions representing the voltages across the resistor and across the capacitor.

a. $770 Ω$ ; b. 0.16 A; c. $I = (0.16 A) cos (120 π t - 0.33 π)$ ; d. $v_{R} = 62 cos (120 π t - 0.33 π)$ ; $v_{C} = 103 cos (120 π t - 0.83 π)$

A resistor and inductor are connected in series across an ac generator. The emf of the generator is given by $v open parenthesis t close parenthesis equals V sub 0 cos ω t,$ where $V sub 0 equals 120 V$ and $ω equals 120 π rad/s;$ also, $R equals 400 Ω$ and $L equals 1.5 H.$ (a) What is the impedance of the circuit? (b) What is the amplitude of the current through the resistor? (c) Write an expression for the current through the resistor. (d) Write expressions representing the voltages across the resistor and across the inductor.

In an RLC series circuit, the voltage amplitude and frequency of the source are 100 V and 500 Hz, respectively, an $R equals 500 Ω,$ $L equals 0.20 H,$ and $C equals 2.0 μ F.$ (a) What is the impedance of the circuit? (b) What is the amplitude of the current from the source? (c) If the emf of the source is given by $v open parenthesis t close parenthesis equals open parenthesis 100 V close parenthesis sin 1000 π t$ , how does the current vary with time? (d) Repeat the calculations with C changed to $0.20 μ F.$

a. $690 Ω$ ; b. 0.15 A; c. $I = (0.15 A) sin (1000 π t - 0.753)$ ; d. $1100 Ω$ , 0.092 A, $I = (0.092 A) sin (1000 π t + 1.09)$

An RLC series circuit with $R equals 600 Ω$ , $L equals 30 mH,$ and $C equals 0.050 μ F$ is driven by an ac source whose frequency and voltage amplitude are 500 Hz and 50 V, respectively. (a) What is the impedance of the circuit? (b) What is the amplitude of the current in the circuit? (c) What is the phase angle between the emf of the source and the current?

For the circuit shown below, what are (a) the total impedance and (b) the phase angle between the current and the emf? (c) Write an expression for $i (t) .$

Figure shows a circuit with a voltage source 170 V, sine 120 pi t, a resistor of 5 ohm, a capacitor of 400 microfarad and an inductor of 25 milihenry all connected in series.

a. $5.7 Ω$ ; b. $29 °$ ; c. $I = (30. A) cos (120 π t + 0.51)$

Adapted from University Physics Volume 2 by OpenStax (openstax.org), licensed under CC BY-NC-SA 4.0. Changes were made. License: CC-BY-NC-SA-4.0.