Series LCR circuit

Plug a resistor, an inductor and a capacitor into the same loop with an AC source, and something interesting happens: the inductor and capacitor fight each other. One pushes the current's phase forward, the other pulls it back, and depending on which wins, the circuit behaves as if it were more resistive, more inductive, or more capacitive than any single part alone. The series LCR circuit is where AC theory earns its keep — everything you learned separately about R, L and C in isolation now has to be combined correctly, using rotating phasors rather than ordinary addition. Get the phasor picture right and impedance, phase angle and current all fall out in a few lines; get it wrong (adding voltages arithmetically, say) and you overshoot the true answer every time. This is also the circuit that sets up resonance, the next topic in the unit.

The core insight

In a series LCR circuit the same current flows through R, L and C, but their voltage phasors add as $V_0=\sqrt{V_R^2+(V_L-V_C)^2}$ rather than arithmetically, giving impedance $Z=\sqrt{R^2+(X_L-X_C)^2}$ and phase angle $\tan\phi=(X_L-X_C)/R$ .

The core idea

Because R, L and C sit in series, exactly the same instantaneous current $i=I_0\sin\omega t$ flows through all three — there is nothing to argue about there. What differs is how each element's own voltage drop relates in phase to that shared current.

$V_R=I_0R$ is in phase with the current — a resistor never introduces a phase shift.
$V_L=I_0X_L$ leads the current by $\pi/2$ (90°), where $X_L=\omega L$ is the inductive reactance.
$V_C=I_0X_C$ lags the current by $\pi/2$ , where $X_C=1/(\omega C)$ is the capacitive reactance.

Building the phasor diagram. Draw the current phasor $I_0$ along a reference (horizontal) axis. $V_R$ is drawn along the same axis, since it is in phase. $V_L$ is drawn rotated 90° ahead of $I_0$ , and $V_C$ is drawn rotated 90° behind — so $V_L$ and $V_C$ lie on the same perpendicular line but point in opposite directions, and only their difference $(V_L-V_C)$ survives. Adding $V_R$ and the net $(V_L-V_C)$ as perpendicular vectors and taking the resultant gives the source voltage:

V_0=\sqrt{V_R^2+(V_L-V_C)^2}=I_0\sqrt{R^2+(X_L-X_C)^2}

where: $V_0$ is the peak source voltage, $V_R$ , $V_L$ , $V_C$ are the peak voltages across each element, $I_0$ is the peak current, and $X_L$ , $X_C$ are the inductive and capacitive reactances (ohms). Factoring out the current defines impedance, the AC circuit's overall opposition to current flow:

Z=\sqrt{R^2+(X_L-X_C)^2}

where: $Z$ is impedance (ohms), $R$ is resistance, and $(X_L-X_C)$ is the net reactance. This gives the AC analogue of Ohm's law, $V_0=I_0Z$ (equivalently $V_{rms}=I_{rms}Z$ ), and fixes the phase angle by which voltage leads current:

\tan\phi=\frac{X_L-X_C}{R}

so that the instantaneous current is $i=I_m\sin(\omega t-\phi)$ , with peak value $I_m=V_0/Z$ . Three regimes follow directly from the sign of $(X_L-X_C)$ : if $X_L>X_C$ the circuit is net inductive ( $\phi>0$ , current lags voltage); if $X_L<X_C$ it is net capacitive ( $\phi<0$ , current leads voltage); if $X_L=X_C$ then $\phi=0$ and $Z=R$ is at its minimum — the condition for resonance.

Figure

Phasor diagram of a series LCR circuit showing current along the reference axis, VR in phase, VL leading by ninety degrees, VC lagging by ninety degrees, and the resultant V0 at angle phi

Derivation AThe impedance of a series LCR circuit

Given resistance $R$ , inductive reactance $X_L$ and capacitive reactance $X_C$ of a series LCR circuit driven at angular frequency $\omega$ , derive the impedance $Z$ from the phasor construction, then read off the phase angle $\phi$ by which the applied voltage leads (or lags) the current, without solving the underlying differential equation.

Step 1. Fix the common current. Since R, L and C are in series, the same current $i=I_0\sin\omega t$ passes through each. Take this as the reference phasor along the horizontal axis of the diagram.

Step 2. Draw each element's voltage phasor relative to that current. $V_R=I_0R$ is in phase (along the reference axis). $V_L=I_0X_L$ leads by $90°$ (drawn straight up). $V_C=I_0X_C$ lags by $90°$ (drawn straight down). $V_L$ and $V_C$ therefore lie on the same vertical line, pointing opposite ways, so their vector sum is simply the difference $(V_L-V_C)$ along that vertical.

Step 3. Add the two remaining perpendicular components, $V_R$ (horizontal) and $(V_L-V_C)$ (vertical), by the Pythagorean rule to get the resultant source voltage: $V_0=\sqrt{V_R^2+(V_L-V_C)^2}$ . Substituting $V_R=I_0R$ , $V_L=I_0X_L$ , $V_C=I_0X_C$ gives $V_0=I_0\sqrt{R^2+(X_L-X_C)^2}$ .

Step 4. Define impedance as the factor relating $V_0$ to $I_0$ , exactly as resistance relates voltage to current in a DC circuit: $Z=V_0/I_0=\sqrt{R^2+(X_L-X_C)^2}$ . The phase angle is the angle of the resultant phasor above the reference axis, $\tan\phi=(V_L-V_C)/V_R=(X_L-X_C)/R$ , since the $I_0$ common to every term cancels. Result: $Z=\sqrt{R^2+(X_L-X_C)^2}$ and $\tan\phi=(X_L-X_C)/R$ , obtained purely from vector (phasor) addition — no calculus needed.

Worked example

A series LCR circuit has $R=30\ \Omega$ , $L=0.4\ \text{H}$ , $C=50\ \mu\text{F}$ , connected to a 200 V (RMS), 50 Hz AC supply. Find the impedance, RMS current, phase angle, and the RMS voltage across the capacitor.

Step 1. Find $\omega$ and the two reactances. $\omega=2\pi f=2\pi\times50=314.16\ \text{rad/s}$ . Then $X_L=\omega L=314.16\times0.4=125.7\ \Omega$ , and $X_C=\dfrac{1}{\omega C}=\dfrac{1}{314.16\times50\times10^{-6}}=63.7\ \Omega$ .

Step 2. Combine into impedance. $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{30^2+(125.7-63.7)^2}=\sqrt{900+3844}=\sqrt{4744}=68.9\ \Omega$ .

Step 3. Find the RMS current using the AC form of Ohm's law. $I_{rms}=\dfrac{V_{rms}}{Z}=\dfrac{200}{68.9}=2.90\ \text{A}$ .

Step 4. Find the phase angle. $\tan\phi=\dfrac{X_L-X_C}{R}=\dfrac{62.0}{30}=2.067\Rightarrow\phi=\tan^{-1}(2.067)=64.2°$ . Since $X_L>X_C$ , the circuit is net inductive, so current lags voltage by $64.2°$ .

Step 5. Find $V_{C,rms}=I_{rms}\times X_C=2.90\times63.7=184.7\ \text{V}$ — close to the 200 V source even though it is only one of three elements, a reminder that individual element voltages need not add simply to the source voltage.

Impedance $Z=68.9\ \Omega$ , RMS current $I_{rms}=2.90\ \text{A}$ , phase angle $\phi=64.2°$ (current lags), and $V_{C,rms}=184.7\ \text{V}$ .

Competitive stretch

Keep $R=40\ \Omega$ but now suppose $X_L=30\ \Omega$ and $X_C=90\ \Omega$ at 100 Hz, with a 100 V RMS source. Here $Z=\sqrt{40^2+(30-90)^2}=\sqrt{1600+3600}=\sqrt{5200}=72.1\ \Omega$ , so $I_{rms}=100/72.1=1.39\ \text{A}$ . Now $\tan\phi=(30-90)/40=-1.5$ , giving $\phi=-56.3°$ : since $X_C>X_L$ the circuit is net capacitive and current leads voltage. The capacitor voltage is $V_{C,rms}=I_{rms}X_C=1.39\times90=125.1\ \text{V}$ — this actually exceeds the 100 V source, which is normal because $V_L$ and $V_C$ partially cancel in the phasor sum rather than adding arithmetically.

Common traps

Adding $V_R$ , $V_L$ , $V_C$ arithmetically instead of combining them as phasors — the correct relation is $V_0=\sqrt{V_R^2+(V_L-V_C)^2}$ , not $V_R+V_L+V_C$ .
Dropping or "fixing" a negative sign in $\tan\phi=(X_L-X_C)/R$ — a negative value correctly means the circuit is capacitive and current leads, it is not an error.
Assuming $V_L$ or $V_C$ individually cannot exceed the source voltage — they routinely can, since they are out of phase and partly cancel in the vector sum.
Reaching for Kirchhoff's law and calculus as the primary solution method — the phasor-diagram method is the prescribed route and is faster for finding $Z$ , $\phi$ and $I$ .

Ask the tutor: if you'd like, ask the tutor to walk through drawing the phasor diagram for a capacitive-dominant LCR circuit step by step, or to connect this result to the resonance condition covered next.

Why do we use the phasor-diagram method instead of just solving Kirchhoff's voltage law directly?

Kirchhoff's law does give the correct governing equation, $L\frac{di}{dt}+iR+\frac{q}{C}=V_0\sin\omega t$ , but solving that differential equation directly is slow. Since every voltage and the current in a series LCR circuit oscillate at the same frequency $\omega$ , differing only in amplitude and phase, they can be represented as rotating phasors and combined with ordinary vector addition instead. That gives the same steady-state result far faster, which is why CISCE prescribes phasors as the primary technique, with Kirchhoff's law used only to justify the setup.

How do I remember which way VL and VC point on the phasor diagram?

Anchor it to the current phasor, drawn along the reference axis. Inductors oppose changes in current, so their voltage phasor is ahead of the current — draw $V_L$ rotated 90 degrees anticlockwise (leading). Capacitors store charge in response to voltage that built up from past current, so their voltage lags — draw $V_C$ rotated 90 degrees clockwise (lagging). Since they point in exactly opposite directions on the same line, you only ever need their difference $(V_L-V_C)$ , never both added separately.

What happens to Z and phi exactly at resonance, where XL equals XC?

When $X_L=X_C$ , the term $(X_L-X_C)$ in the impedance formula becomes zero, so $Z=\sqrt{R^2+0}=R$ — the impedance is at its minimum possible value for that circuit, equal to just the resistance. Correspondingly $\tan\phi=(X_L-X_C)/R=0$ , so $\phi=0$ and the current is exactly in phase with the applied voltage, behaving as if only a resistor were present. This special frequency is the resonant frequency, and it is covered as its own topic building directly on this result.

Series LCR circuit

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads