32.5 The Series $R L C$ Circuit

The circuit of FIGURE 32.16, where a resistor, inductor, and capacitor are in series, is called a series $R L C$ circuit. The series $R L C$ circuit has many important applications because, as you will see, it exhibits resonance behavior.

The figure illustrates a series R L C circuit.

FIGURE 32.16 Full Alternative Text

The analysis, which is very similar to our analysis of the $R C$ circuit in Section 32.3, will be based on a phasor diagram. Notice that the three circuit elements are in series with each other and, together, are in parallel with the emf. We can draw two conclusions that form the basis of our analysis:

The instantaneous current of all three elements is the same: $i = i_{R} = i_{L} = i_{C}$ .
The sum of the instantaneous voltages matches the emf: $ε = v_{R} + v_{L} + v_{C}$ .

Using phasors to analyze an $R L C$ 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 $i$ .

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The current and voltage of a resistor are in phase, so draw a resistor voltage phasor parallel to the current phasor $I$ . The capacitor current leads the capacitor voltage by $90 °$ , so draw a capacitor voltage phasor that is $90 °$ behind the current phasor. The inductor current lags the voltage by $90 °$ , so draw an inductor voltage phasor $90 °$ ahead of the current phasor.

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The instantaneous voltages satisfy $ε = v_{R} + v_{L} + v_{C}$ . In terms of phasors, this is a vector addition. We can do the addition in two steps. Because the capacitor and inductor phasors are in opposite directions, their vector sum has length $V_{L} - V_{C}$ . Adding the resistor phasor, at right angles, then gives the emf phasor $ε$ at angle $ω t$ .

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The length $ε_{0}$ of the emf phasor is the hypotenuse of a right triangle. Thus $ε_{ 0} ^{2} = V_{ R} ^{2} + {(V_{ L} - V_{ C})}^{2}$

If $V_{L} > V_{C}$ , which we’ve assumed, then the instantaneous current $i$ lags the emf by a phase angle $ϕ$ . We can write the current, in terms of $ϕ$ , as

i = I cos (ω t - ϕ)

(32.22)

Of course, there’s no guarantee that $V_{ L}$ will be larger than $V_{ C}$ . If the opposite is true, $V_{L} < V_{C}$ , the emf phasor is on the other side of the current phasor. Our analysis is still valid if we consider $ϕ$ to be negative when $i$ is ccw from $ε$ . Thus $ϕ$ can be anywhere between $- 90 °$ and $+ 90 °$ .

Now we can continue much as we did with the $R C$ circuit. Based on the right triangle, ${ε_{0}}^{2}$ is

ε_{​ 0} ​^{2} = V_{​ ​ R} ​^{2} + {(V_{​ ​ L} - V_{​ ​ C})}^{2} = R^{2} + {(X_{​ L} - X_{​ C})}^{2} I^{2}

(32.23)

where we wrote each of the peak voltages in terms of the peak current $I$ and a resistance or a reactance. Consequently, the peak current in the $R L C$ circuit is

I = \frac{ε_{0}}{\sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}} = \frac{ε_{0}}{\sqrt{R^{2} + {(ω L - 1 / ω C)}^{2}}}

(32.24)

The three peak voltages are then found from $V_{R} = I R, V_{L} = I X_{L}$ , and $V_{C} = I X_{C}$ .