The series LCR circuit
Consider a resistor $R$, an inductor $L$ and a capacitor $C$ connected in series across an ac source whose voltage is $v = v_m \sin\omega t$. Because the three elements lie in a single loop, the same current flows through each of them at every instant, with one common amplitude and phase. Writing this current as $i = i_m \sin(\omega t + \phi)$, the task is to find the current amplitude $i_m$ and the phase angle $\phi$ between the current and the source voltage.
Applying Kirchhoff's loop rule to the circuit gives the relation between the instantaneous voltages across the three elements and the source:
$$ v_L + v_R + v_C = v $$NCERT solves this in two ways: by the technique of phasors and by analysing the loop equation directly. The phasor approach is the one tested at NEET, so it is developed below.
A series LCR circuit. The same instantaneous current $i$ passes through $R$, $L$ and $C$.
Impedance and phase angle
Take the current phasor $\mathbf{I}$ as the reference. The voltage across the resistor is in phase with the current; the voltage across the inductor leads the current by $\pi/2$; and the voltage across the capacitor lags the current by $\pi/2$. Their amplitudes are $v_{Rm}=i_m R$, $v_{Lm}=i_m X_L$ and $v_{Cm}=i_m X_C$, where $X_L=\omega L$ and $X_C=1/\omega C$ are the inductive and capacitive reactances.
Since $\mathbf{V_L}$ and $\mathbf{V_C}$ point along the same line in opposite directions, they combine into a single phasor of magnitude $|v_{Cm}-v_{Lm}|$. The source phasor $\mathbf{V}$ is the hypotenuse of a right triangle whose perpendicular sides are $v_{Rm}$ and $(v_{Cm}-v_{Lm})$. Applying the Pythagoras theorem and dividing by $i_m$ gives the impedance:
$$ Z = \sqrt{R^2 + (X_L - X_C)^2}, \qquad i_m = \frac{v_m}{Z} $$The phase angle between the source voltage and current follows from the same triangle:
$$ \tan\phi = \frac{X_C - X_L}{R} $$| Condition | Reactance balance | Phase behaviour | Circuit nature |
|---|---|---|---|
X_C > X_L | capacitive term dominates | current leads voltage | predominantly capacitive |
X_C < X_L | inductive term dominates | current lags voltage | predominantly inductive |
X_C = X_L | reactances cancel | current in phase, $\phi=0$ | resistive (resonance) |
The impedance triangle
Dividing every side of the voltage phasor triangle by the common current $i_m$ converts it into the impedance triangle. Its base is $R$, its perpendicular is $(X_L - X_C)$, and its hypotenuse is the impedance $Z$. The angle between $R$ and $Z$ is the phase angle $\phi$.
The impedance triangle. Resistance forms the base, net reactance the height, and impedance the hypotenuse.
The whole LCR analysis rests on adding rotating vectors. Revisit Phasors if the phasor-triangle step feels shaky.
Resonance in the series LCR circuit
For a driven RLC circuit the current amplitude is $i_m = v_m/Z$, with $X_C = 1/\omega C$ and $X_L = \omega L$. As the source frequency $\omega$ is varied, there is one particular frequency $\omega_0$ at which the inductive and capacitive reactances become equal, $X_L = X_C$. The reactive term then vanishes, the impedance drops to its minimum value $Z = R$, and the current amplitude rises to its maximum $i_m = v_m/R$. This is resonance.
Setting $\omega_0 L = 1/\omega_0 C$ gives the resonant angular frequency and the corresponding resonant frequency:
$$ \omega_0 = \frac{1}{\sqrt{LC}}, \qquad \nu_0 = \frac{1}{2\pi\sqrt{LC}} $$At resonance the source voltage and current are in phase, so $\phi=0$ and the power factor $\cos\phi = 1$. This is the principle behind tuning a radio: the capacitance of the tuning circuit is varied until $\omega_0$ matches the broadcast frequency, maximising the current for that station.
Series resonance means minimum Z, maximum current
A frequent mistake is to claim the impedance is maximum at resonance. In a series LCR circuit the opposite holds: at $\omega_0$, $X_L=X_C$, the reactive term cancels, so $Z=R$ is minimum and the current is maximum. (Maximum impedance at resonance is the behaviour of a parallel LC circuit, not the series one.)
At series resonance: $X_L = X_C$, $Z_{\min}=R$, $i_{\max}=v_m/R$, $\phi=0$, $\cos\phi=1$. Resonance needs both $L$ and $C$ — never an RL or RC circuit.
Current versus frequency. A smaller resistance gives a taller, sharper peak; the half-power points define the bandwidth $2\Delta\omega$.
NCERT illustrates this with $L = 1.00$ mH and $C = 1.00$ nF, for which $\omega_0 = 1/\sqrt{LC} = 1.00\times10^6$ rad/s. With source amplitude $v_m = 100$ V, the curve for $R = 100\,\Omega$ peaks twice as high as the curve for $R = 200\,\Omega$, since $i_m = v_m/R$ at resonance. Resonance appears only when both $L$ and $C$ are present, because only then can the inductor and capacitor voltages — being $180^\circ$ out of phase — cancel, leaving the full source voltage across $R$.
Quality factor and sharpness
How sharply the current peaks near $\omega_0$ is measured by the quality factor $Q$. NCERT gives, for a series RLC circuit,
$$ Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R} $$A high $Q$ corresponds to a narrow, tall resonance curve — a circuit that responds strongly to a small band of frequencies around $\omega_0$ and is therefore highly selective. The half-power frequencies, where the power falls to half its resonant value (current to $i_{\max}/\sqrt2$), lie symmetrically at $\omega_0 \pm R/2L$. Their separation defines the bandwidth:
$$ 2\Delta\omega = \frac{R}{L}, \qquad Q = \frac{\omega_0}{2\Delta\omega} $$| Quantity | Expression | Effect of increasing R |
|---|---|---|
| Resonant frequency | ω₀ = 1/√(LC) | unchanged |
| Impedance at resonance | Z = R | increases (lower peak current) |
| Quality factor | Q = ω₀L/R = 1/(ω₀CR) | decreases |
| Bandwidth | 2Δω = R/L | increases (broader curve) |
Increasing $R$ leaves $\omega_0$ untouched but lowers $Q$ and widens the band, so the resonance becomes less sharp. Decreasing $R$ raises $Q$, narrows the band and sharpens the peak. Bandwidth and quality factor are therefore inversely related.
Worked examples
A resistor of $200\,\Omega$ and a capacitor of $15.0\,\mu\text{F}$ are connected in series to a 220 V, 50 Hz ac source. Find the current, and the rms voltages across the resistor and capacitor.
The capacitive reactance is $X_C = 1/(2\pi\nu C) \approx 212.3\,\Omega$, so the impedance is $Z = \sqrt{R^2 + X_C^2} = \sqrt{200^2 + 212.3^2} \approx 291.5\,\Omega$. The current is $I = V/Z = 220/291.5 \approx 0.755$ A. Then $V_R = IR = 151$ V and $V_C = IX_C = 160.3$ V. Their algebraic sum, 311.3 V, exceeds the 220 V source — the paradox resolves because the two voltages are $90^\circ$ out of phase: $\sqrt{V_R^2 + V_C^2} = 220$ V, the source voltage.
A series LCR circuit has $L = 10$ mH and $C = 1\,\mu\text{F}$. Find the resonant frequency.
$\nu_0 = \dfrac{1}{2\pi\sqrt{LC}} = \dfrac{1}{2\pi\sqrt{(10\times10^{-3})(1\times10^{-6})}} \approx 1.59\times10^{3}\ \text{Hz} = 1.59$ kHz. The value of $R$ does not enter the resonant frequency. This is precisely NEET 2023 Q.26.
Series LCR circuit and resonance in one screen
- Impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2}$, with current amplitude $i_m = v_m/Z$.
- Phase angle: $\tan\phi = (X_C - X_L)/R$; current leads if $X_C>X_L$, lags if $X_C<X_L$.
- Resonance occurs when $X_L = X_C$, giving $\omega_0 = 1/\sqrt{LC}$ and $\nu_0 = 1/(2\pi\sqrt{LC})$.
- At resonance $Z = R$ (minimum), current is maximum, $\phi = 0$, power factor $= 1$.
- Quality factor $Q = \omega_0 L/R = 1/(\omega_0 C R)$; higher $Q$ means sharper resonance and smaller bandwidth $2\Delta\omega = R/L$.
- Resonance needs both $L$ and $C$ — it cannot occur in an RL or RC circuit.