R, L, C series circuits

Plug a resistor, an inductor, and a capacitor in series and drive them from a wall socket, and the current that flows is still a clean sine wave at the supply frequency — but it no longer peaks at the same instant as the voltage. A resistor lets voltage and current move in lockstep; an inductor makes the voltage lead; a capacitor makes the current lead. Put all three in one loop and the circuit settles into a single net behaviour — inductive, capacitive, or purely resistive — decided by which reactance wins. This article builds the tool that predicts that outcome: impedance, the AC circuit's single combined opposition to current, together with the phase angle that tells you exactly how far current and voltage have drifted apart. It is the analytical core of the "Single-Phase AC Circuits" unit, and everything that follows — phasor diagrams, the $j$ -operator, power factor — is built on it.

The core insight

In a series R-L-C circuit the impedance $Z = R + j(X_L - X_C)$ sets both the current magnitude $I = V/|Z|$ and the phase angle between current and voltage, with the sign of $X_L - X_C$ deciding whether the circuit behaves as inductive or capacitive.

The core idea

A resistor obeys $v = Ri$ with voltage and current always in step. An inductor obeys $v = L\,di/dt$ ; for a sinusoidal current this makes the voltage lead the current by $90^\circ$ . A capacitor obeys $i = C\,dv/dt$ , which makes the current lead the voltage by $90^\circ$ . Each of these quarter-cycle shifts is the signature of energy sloshing back into a magnetic or electric field rather than being dissipated, and it is the reason L and C cannot simply be added to R like ordinary resistances. Each element's AC opposition to current, measured in ohms, is:

X_L = \omega L, \qquad X_C = \frac{1}{\omega C}

where: $X_L$ is inductive reactance, $X_C$ is capacitive reactance, $\omega = 2\pi f$ is the angular frequency (rad/s), $L$ is inductance (H), $C$ is capacitance (F), and $f$ is the supply frequency (Hz). $X_L$ grows with frequency — a fast-changing current is opposed more strongly by an inductor — while $X_C$ shrinks with frequency, since a capacitor charges and discharges more freely as the drive speeds up. Because $X_L$ and $X_C$ act $180^\circ$ apart in phase, they subtract rather than add when combined with R to form the impedance:

Z = R + jX, \qquad X = X_L - X_C, \qquad |Z| = \sqrt{R^2 + (X_L - X_C)^2}, \qquad \theta = \tan^{-1}\!\left(\frac{X_L - X_C}{R}\right)

where: $Z$ is the impedance (a complex quantity with magnitude and phase), $X$ is the net reactance, $|Z|$ is impedance magnitude (ohms), and $\theta$ is the phase angle by which current lags ( $\theta > 0$ ) or leads ( $\theta < 0$ ) the voltage. The $j$ -operator's full algebra is developed in the next article; here it is enough to picture $R$ and $X$ as two perpendicular sides of a right triangle whose hypotenuse is $Z$ , as shown below.

Figure

Impedance triangle with resistance along the base capacitive or inductive reactance along the height and impedance as the hypotenuse at angle theta

Three limiting cases are worth fixing directly: pure R gives $Z = R$ and $\theta = 0$ ; pure L gives $Z = jX_L$ and $\theta = 90^\circ$ (current lags); pure C gives $Z = -jX_C$ and $\theta = -90^\circ$ (current leads). Ohm's law for AC, using RMS phasor quantities, is $V = IZ$ , so current magnitude follows from $I = V/|Z|$ once $|Z|$ is known.

Derivation AImpedance magnitude and phase of a series R-L-C circuit

The goal is to combine a resistor's resistance with an inductor's and a capacitor's reactances — three elements governed by three different voltage-current relationships — into one impedance magnitude and one phase angle that together let $I = V/|Z|$ and the lag or lead of current behind voltage be read off directly, without carrying separate equations for each element through every problem.

Step 1. Write each element's voltage for a common series current $i(t) = I_m\sin(\omega t)$ . The resistor gives $v_R = Ri$ , in phase with $i$ . The inductor gives $v_L = L\,di/dt$ , which works out to a sinusoid of amplitude $\omega L I_m$ leading $i$ by $90^\circ$ — so its opposition, $X_L = \omega L$ , acts along a direction rotated $+90^\circ$ from R. The capacitor gives $i = C\,dv_C/dt$ , which inverts to a sinusoid of amplitude $I_m/(\omega C)$ lagging $i$ by $90^\circ$ — so $X_C = 1/(\omega C)$ acts along $-90^\circ$ from R.

Step 2. Add the three voltage contributions as phasors, not as ordinary numbers, because they point in different directions: $R$ along $0^\circ$ , $X_L$ along $+90^\circ$ , $X_C$ along $-90^\circ$ . Since $X_L$ and $X_C$ point in exactly opposite directions, they partially cancel; only their difference, $X = X_L - X_C$ , survives as the net reactive part.

Step 3. Treat $R$ and $X$ as the two perpendicular legs of a right triangle whose hypotenuse is the impedance magnitude. By Pythagoras, $|Z| = \sqrt{R^2 + X^2} = \sqrt{R^2 + (X_L - X_C)^2}$ . This single number converts the RMS supply voltage directly into RMS current through $I = V/|Z|$ .

Step 4. Read the phase angle from the same triangle: $\theta = \tan^{-1}(X/R) = \tan^{-1}\!\left(\dfrac{X_L - X_C}{R}\right)$ . A positive $\theta$ means $X_L > X_C$ , the circuit is net inductive, and current lags voltage by $\theta$ ; a negative $\theta$ means $X_C > X_L$ , the circuit is net capacitive, and current leads voltage by $|\theta|$ . Result: $|Z| = \sqrt{R^2 + (X_L - X_C)^2}$ , $\theta = \tan^{-1}\!\left(\dfrac{X_L - X_C}{R}\right)$ , $I = V/|Z|$ — the working formula for any series R-L-C combination.

Worked example

A series R-L-C circuit has $R = 10\ \Omega$ , $L = 0.15\ \text{H}$ , and $C = 60\ \mu\text{F}$ , connected to a 100 V, 50 Hz supply. Find the net reactance, impedance, current, and state whether the circuit is inductive or capacitive.

Step 1. Angular frequency: $\omega = 2\pi f = 2\pi \times 50 = 314.16\ \text{rad/s}$ . Inductive reactance: $X_L = \omega L = 314.16 \times 0.15 = 47.12\ \Omega$ .

Step 2. Capacitive reactance: $X_C = \dfrac{1}{\omega C} = \dfrac{1}{314.16 \times 60\times10^{-6}} = \dfrac{1}{0.01885} = 53.05\ \Omega$ .

Step 3. Net reactance: $X = X_L - X_C = 47.12 - 53.05 = -5.93\ \Omega$ . The negative sign shows $X_C$ dominates, so the circuit is net capacitive.

Step 4. Impedance magnitude: $|Z| = \sqrt{R^2 + X^2} = \sqrt{10^2 + (-5.93)^2} = \sqrt{100 + 35.2} = \sqrt{135.2} = 11.63\ \Omega$ .

Step 5. Current and phase: $I = \dfrac{100}{11.63} = 8.60\ \text{A}$ . $\theta = \tan^{-1}\!\left(\dfrac{-5.93}{10}\right) = -30.6^\circ$ , so current leads voltage by $30.6^\circ$ , consistent with the net capacitive nature. Answer: $X = -5.93\ \Omega$ , $|Z| = 11.63\ \Omega$ , $I = 8.60\ \text{A}$ , current leads by $30.6^\circ$ .

Competitive stretch

Find the voltage drop magnitude across each element and check they do not simply add to 100 V. Step 1. $V_R = IR = 8.60 \times 10 = 86.0\ \text{V}$ . Step 2. $V_L = IX_L = 8.60 \times 47.12 = 405.2\ \text{V}$ . Step 3. $V_C = IX_C = 8.60 \times 53.05 = 456.2\ \text{V}$ . Step 4. These are far larger than the 100 V supply individually, yet they combine as phasors: $V = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{86.0^2 + (405.2-456.2)^2} = \sqrt{7396 + 2601} = \sqrt{9997} \approx 100\ \text{V}$ , matching the supply. The lesson: element voltage drops in a series R-L-C circuit can individually exceed the supply voltage — they only combine correctly as phasors, never by plain arithmetic addition.

Common traps

Adding $X_L$ and $X_C$ arithmetically instead of subtracting — they act $180^\circ$ apart in phase, so the net reactance is $X_L - X_C$ , never $X_L + X_C$ .
Forgetting to convert $L$ from mH to H or $C$ from $\mu\text{F}$ to F before computing $\omega L$ or $1/(\omega C)$ — a very common arithmetic slip that throws off every later step.
Assuming the circuit is always inductive; when $X_C > X_L$ the circuit is capacitive and current leads voltage — always check the sign of $X = X_L - X_C$ before stating the phase relationship.
Adding individual element voltage drops $V_R, V_L, V_C$ arithmetically to check against the supply voltage — they must be combined as phasors, and $V_L, V_C$ can each individually exceed the supply voltage.

Ask the tutor: if a problem gives you power dissipated or a voltage-drop reading instead of R, L, C values directly, ask the AI tutor to show how to work backwards to impedance and reactance.

Why do X_L and X_C subtract instead of add when they are both a kind of resistance

They are not ordinary resistances — they are reactances, and each pushes the voltage a quarter cycle out of phase with current, but in opposite directions. An inductor's voltage leads current by 90 degrees while a capacitor's voltage lags current by 90 degrees, so on a phasor diagram they point in exactly opposite directions along the same axis. Combining opposite phasors means subtracting their magnitudes, giving the net reactance X = X_L minus X_C, not the sum. Only when one of them is absent does the other stand alone.

How can the voltage across the inductor or capacitor be larger than the supply voltage

Because the element voltages are out of phase with each other, not in phase, they never add by simple arithmetic. V_L and V_C point in opposite directions on the phasor diagram, so a large V_L and a nearly-as-large V_C can mostly cancel each other, leaving only a modest resultant that combines with V_R to match the supply. Individually either one can be several times the supply voltage — this is the everyday behaviour behind series resonance, and it is real, not a calculation error, so always verify by phasor addition, never plain addition.

How do I quickly tell if a series R-L-C circuit will be inductive or capacitive without a calculator

Compare X_L = omega L with X_C = 1 over (omega C) at the given frequency. If X_L is the bigger number the circuit is net inductive and current lags voltage; if X_C is bigger it is net capacitive and current leads voltage. As a rule of thumb, low frequency favours capacitive behaviour because X_C grows as frequency falls while X_L shrinks, and high frequency favours inductive behaviour for the opposite reason. Compute both reactances first, then subtract, then worry about the impedance magnitude and angle.

R, L, C series circuits

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads