Derivation

AC Circuits: Phasors, Impedance and Series RLC Resonance

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Statement

For a linear series RLC circuit driven by a sinusoidal voltage source \(V(t)=V_0 \cos(\omega t)\), the steady-state current is \(I(t)=I_0 \cos(\omega t - \phi)\) where the amplitude \(I_0\) and phase lag \(\phi\) are determined by the complex impedance \(Z = R + i\left(\omega L - \frac{1}{\omega C}\right)\). Specifically, \(I_0 = \frac{V_0}{|Z|}\) and \(\tan\phi = \frac{\omega L - 1/(\omega C)}{R}\). The system resonates at \(\omega_0 = \frac{1}{\sqrt{LC}}\), where the impedance is purely resistive (\(Z=R\)) and the current amplitude is maximal.

Why it matters

AC circuit analysis is the foundation of power transmission, radio communication, and signal processing. The phasor method transforms differential equations describing inductors and capacitors into algebraic equations involving complex numbers. This allows engineers to analyze circuits of arbitrary complexity using the same series-parallel reduction rules as DC resistors, while retaining the frequency-dependent phase relationships that determine power flow and resonance.

Resonance in RLC circuits is the electrical analogue of mechanical resonance. It is exploited in tuning circuits to select specific frequencies from a spectrum and is a critical design consideration in power systems, where resonant conditions can cause dangerous voltage or current magnification.

Assumptions

Linearity of components.The resistance \(R\), inductance \(L\), and capacitance \(C\) must be constant, independent of voltage and current. If \(R\), \(L\), or \(C\) vary with the signal (e.g., a diode or a nonlinear inductor), superposition fails and the phasor method is invalid; one must solve the nonlinear differential equations directly.

Steady-state sinusoidal drive.The analysis assumes the source has been operating for a long time, so that transient terms (exponentials decaying as \(e^{-t/\tau}\)) have vanished. During transients, the current is not purely sinusoidal, and the impedance concept does not apply.

Single frequency \(\omega\).The source must be a pure sinusoid. If the source contains multiple frequencies (a square wave, for instance), each frequency component must be analyzed separately using its own impedance, and the results summed (superposition), provided the circuit is linear.

Quasistatic approximation.The physical dimensions of the circuit elements must be small compared to the wavelength \(\lambda = c/\omega\) of the signal. If the wire lengths are comparable to \(\lambda\), the circuit becomes a transmission line, and lumped-element impedance \(Z_L = i\omega L\) is no longer valid; distributed effects and radiation must be considered.

Derivation

\[ V(t) = V_0 \cos(\omega t) = \mathrm{Re}\left[ V_0 e^{i\omega t} \right] \]

Represent the real sinusoidal voltage as the real part of a complex exponential. This allows differentiation to become multiplication by \(i\omega\). A

\[ V_R = I R, \quad V_L = L \frac{dI}{dt}, \quad I = C \frac{dV_C}{dt} \]

The constitutive laws for the resistor, inductor, and capacitor in the time domain. A

\[ \frac{d}{dt} \left( e^{i\omega t} \right) = i\omega e^{i\omega t} \]

Differentiation of the complex exponential. In the phasor domain, the operator \(d/dt\) is replaced by \(i\omega\). A

\[ \tilde{V}_L = L (i\omega) \tilde{I} = (i\omega L) \tilde{I} \]

Substitute the phasor current \(\tilde{I}\) and the derivative rule into the inductor law. The term \(i\omega L\) is the inductive impedance. B

\[ \tilde{I} = C \frac{d\tilde{V}_C}{dt} = C (i\omega) \tilde{V}_C \implies \tilde{V}_C = \frac{1}{i\omega C} \tilde{I} \]

Substitute the phasor voltage and derivative rule into the capacitor law. The term \(1/(i\omega C)\) is the capacitive impedance. B

\[ Z_R = R, \quad Z_L = i\omega L, \quad Z_C = \frac{1}{i\omega C} = -i\frac{1}{\omega C} \]

Define the complex impedance \(Z\) for each element as the ratio \(\tilde{V}/\tilde{I}\). Note that \(1/i = -i\), so the capacitive impedance is purely imaginary and negative. C

\[ Z_{\text{total}} = Z_R + Z_L + Z_C = R + i\omega L - i\frac{1}{\omega C} = R + i\left(\omega L - \frac{1}{\omega C}\right) \]

For a series circuit, impedances add algebraically, just as resistances do in DC. A

\[ \tilde{I} = \frac{\tilde{V}}{Z_{\text{total}}} = \frac{V_0}{R + i\left(\omega L - \frac{1}{\omega C}\right)} \]

Apply Ohm's law in the phasor domain. The current phasor is the voltage phasor divided by the total impedance. A

\[ |Z| = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} \]

The magnitude of the complex impedance is the modulus of the complex number \(R + iX\), where \(X = \omega L - 1/(\omega C)\) is the reactance. A

\[ I_0 = \frac{V_0}{|Z|} = \frac{V_0}{\sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}} \]

The amplitude of the current is the ratio of the voltage amplitude to the impedance magnitude. A

\[ \phi = \arg(Z) = \arctan\left( \frac{\omega L - 1/(\omega C)}{R} \right) \]

The phase angle of the impedance is the angle of the complex number in the Argand plane. The current lags the voltage by this angle \(\phi\) (if \(X>0\), inductive). B

\[ \omega_0 = \frac{1}{\sqrt{LC}} \implies \omega_0 L = \frac{1}{\omega_0 C} \implies Z(\omega_0) = R \]

Resonance occurs when the inductive and capacitive reactances cancel. At this frequency, the impedance is purely real and minimal, so the current amplitude is maximal. C

Result

\[ I_0 = \frac{V_0}{\sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}}, \quad \tan\phi = \frac{\omega L - \frac{1}{\omega C}}{R}, \quad \omega_0 = \frac{1}{\sqrt{LC}} \]

Reading. The current amplitude is maximized at the resonant frequency \(\omega_0\). The phase \(\phi\) indicates whether the circuit is inductive (\(\phi>0\), current lags voltage) or capacitive (\(\phi<0\), current leads voltage). At resonance, \(\phi=0\) and the circuit behaves as a pure resistor.

Units check. \([L] = \mathrm{H = V\cdot s/A}\), \([C] = \mathrm{F = A\cdot s/V}\). \(\sqrt{LC}\) has units \(\mathrm{s}\), so \(1/\sqrt{LC}\) has units \(\mathrm{s^{-1} = rad/s}\). \(\omega L\) has units \(\mathrm{(rad/s)(V\cdot s/A) = V/A = \Omega}\). \(1/(\omega C)\) has units \(\mathrm{1/((rad/s)(A\cdot s/V)) = V/A = \Omega}\). All terms in \(Z\) are in ohms. Consistent.

Limiting cases

DC limit (\(\omega \to 0\)). \(Z_L \to 0\) (short circuit), \(Z_C \to \infty\) (open circuit). The capacitor blocks DC, and the inductor acts as a wire. The series impedance \(|Z|\to\infty\), so no steady current flows: a series RLC circuit passes no DC.
High-frequency limit (\(\omega \to \infty\)). \(Z_L \to \infty\) (open circuit), \(Z_C \to 0\) (short circuit). The inductor blocks high frequencies and the capacitor acts as a wire; again \(|Z|\to\infty\) and \(I_0\to 0\). The current is appreciable only in a band around \(\omega_0\).
Purely resistive (\(L\to 0\), \(C\to\infty\), i.e. the capacitor replaced by a wire). \(Z=R\), \(\phi=0\). The circuit reduces to a simple DC resistor with AC drive; current and voltage are in phase.
Ideal transformer side note. While not part of the series RLC model, Faraday's law gives the transformer ratio \(V_s/V_p = N_s/N_p\). It is the limit of two coupled coils with perfect flux linkage (every turn threads the same flux \(\Phi\), so \(V = N\,d\Phi/dt\) for each coil) and negligible resistance and magnetising current; then \(V_pI_p = V_sI_s\) as well.

Breaks when

Non-sinusoidal drive. If the source is a square wave or pulse, the steady-state current is not a sinusoid. The phasor method applies only to each Fourier component separately, and the total response is the sum of sinusoidal responses at different frequencies.
Non-linear components. If \(R\), \(L\), or \(C\) depend on \(V\) or \(I\) (e.g., a diode, a saturable inductor), the superposition principle fails. The impedance is not a constant, and the phasor algebra is invalid.
Radiation and transmission lines. If the circuit dimensions are comparable to the wavelength \(\lambda = c/\omega\), the lumped-element assumption breaks down. Energy radiates from the wires, and the voltage and current are no longer uniform across the elements; transmission line theory is required.

Failure modes

Confusing \(i\) and \(j\). Mathematicians use \(i\) for the imaginary unit; engineers use \(j\) because \(i\) denotes current. Mixing them leads to confusion in notation, though the physics is identical.
Ignoring the phase angle. Calculating \(I_0 = V_0/|Z|\) but forgetting that the current is shifted by \(\phi\). This leads to errors in power calculations and waveform reconstruction.
Using DC rules for AC. Treating an inductor as a short circuit and a capacitor as an open circuit at all frequencies. This is only true in the DC limit (\(\omega=0\)). At finite \(\omega\), their impedances are frequency-dependent.
Assuming resonance means \(V_L = V_C = 0\). At resonance, \(V_L\) and \(V_C\) are not zero; they are equal in magnitude and opposite in phase, so their sum is zero. The individual voltages can be much larger than the source voltage (voltage magnification).

Discussion

The phasor method is a powerful tool because it converts differential equations into algebraic equations. The key insight is that sinusoidal functions are eigenfunctions of linear time-invariant systems. The response to a sinusoidal input is always a sinusoid of the same frequency, with only the amplitude and phase changed. This allows us to represent the entire time-domain behavior with a single complex number (the phasor) at each frequency.

Resonance is a phenomenon where the energy stored in the inductor and capacitor oscillates back and forth, with the resistor dissipating a small amount each cycle. At resonance, the reactive components cancel, and the circuit appears purely resistive. This is why RLC circuits are used as filters: they can selectively amplify or attenuate specific frequency bands.

The quality factor \(Q = \omega_0 L / R\) measures the sharpness of the resonance. A high \(Q\) circuit has a narrow bandwidth and large voltage magnification across the reactive components. A low \(Q\) circuit has a broad bandwidth and less magnification. The bandwidth \(\Delta\omega = R/L\) is the range of frequencies over which the power is at least half the maximum value. Power itself is set by RMS quantities: with \(V_{\text{rms}} = V_0/\sqrt2\) and \(I_{\text{rms}} = I_0/\sqrt2\), the time-averaged power delivered is \(\langle P\rangle = V_{\text{rms}}I_{\text{rms}}\cos\phi = I_{\text{rms}}^2R\), and \(\cos\phi\) is called the power factor; the inductor and capacitor absorb no net power over a cycle.

In the context of electromagnetic theory, the impedance of a circuit is a manifestation of the boundary conditions on the electromagnetic fields. The inductive and capacitive terms arise from the magnetic and electric field energy stored in the space around the components. The phasor method is a low-frequency approximation of Maxwell's equations, valid when the circuit dimensions are small compared to the wavelength. At higher frequencies, the full wave nature of the fields must be considered, and the concept of a lumped impedance becomes ambiguous.

Common misconceptions. Students often think that the inductor and capacitor "store" energy in the same way a battery stores chemical energy. In reality, they store energy in the electromagnetic fields, and this energy is exchanged with the source and the other reactive component. The resistor is the only component that dissipates energy as heat. Another misconception is that the current and voltage are always in phase in a series circuit; this is only true at resonance or in a purely resistive circuit.

Worked examples

Example 1. A series RLC circuit has \(R = 10\ \Omega\), \(L = 0.1\ \mathrm{H}\), and \(C = 100\ \mu\mathrm{F}\). It is driven by a \(100\ \mathrm{V}\) (peak) sinusoidal source at \(f = 50\ \mathrm{Hz}\). Find the current amplitude, phase angle, and resonant frequency.

\[ \omega = 2\pi f = 2\pi(50) = 100\pi \approx 314.2\ \mathrm{rad/s} \]

Convert frequency to angular frequency. A

\[ X_L = \omega L = 314.2(0.1) = 31.42\ \Omega \]

Calculate the inductive reactance. A

\[ X_C = \frac{1}{\omega C} = \frac{1}{314.2(100 \times 10^{-6})} = \frac{1}{0.03142} \approx 31.83\ \Omega \]

Calculate the capacitive reactance. A

\[ X = X_L - X_C = 31.42 - 31.83 = -0.41\ \Omega \]

Net reactance is slightly capacitive. A

\[ |Z| = \sqrt{R^2 + X^2} = \sqrt{10^2 + (-0.41)^2} = \sqrt{100 + 0.17} \approx 10.01\ \Omega \]

Calculate the impedance magnitude. A

\[ I_0 = \frac{V_0}{|Z|} = \frac{100}{10.01} \approx 9.99\ \mathrm{A} \]

Calculate the current amplitude. A

\[ \phi = \arctan\left(\frac{X}{R}\right) = \arctan\left(\frac{-0.41}{10}\right) \approx -2.4^\circ \]

Calculate the phase angle. Negative means current leads voltage (capacitive). A

\[ \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1 \times 100 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-5}}} = \frac{1}{0.00316} \approx 316.2\ \mathrm{rad/s} \]

Calculate the resonant angular frequency. A

\[ I_0 \approx 9.99\ \mathrm{A}, \quad \phi \approx -2.4^\circ, \quad \omega_0 \approx 316.2\ \mathrm{rad/s} \]

Reading. The circuit is very close to resonance, so the impedance is nearly purely resistive and the current is large. The small negative phase angle indicates a slight capacitive character.

Units check. \(I_0\) is in amperes, \(\phi\) is in degrees, \(\omega_0\) is in rad/s. All consistent with SI units.

Example 2. A series RLC circuit has \(R = 50\ \Omega\), \(L = 0.5\ \mathrm{H}\), and \(C = 20\ \mu\mathrm{F}\). Find the resonant frequency, the quality factor \(Q\), and the bandwidth \(\Delta\omega\).

\[ \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.5 \times 20 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-5}}} \approx 316.2\ \mathrm{rad/s} \]

Calculate the resonant angular frequency. A

\[ Q = \frac{\omega_0 L}{R} = \frac{316.2 \times 0.5}{50} = \frac{158.1}{50} \approx 3.16 \]

Calculate the quality factor. A

\[ \Delta\omega = \frac{R}{L} = \frac{50}{0.5} = 100\ \mathrm{rad/s} \]

Calculate the bandwidth. A

\[ \text{Check: } \frac{\omega_0}{Q} = \frac{316.2}{3.16} \approx 100\ \mathrm{rad/s} \]

Verify that \(\Delta\omega = \omega_0/Q\). A

\[ \omega_0 \approx 316.2\ \mathrm{rad/s}, \quad Q \approx 3.16, \quad \Delta\omega = 100\ \mathrm{rad/s} \]

Reading. The circuit has a moderate quality factor, meaning the resonance is not extremely sharp. The bandwidth is 100 rad/s, so the circuit responds significantly to frequencies within \(\pm 50\) rad/s of the resonant frequency.

Units check. \(\omega_0\) and \(\Delta\omega\) are in rad/s, \(Q\) is dimensionless. Consistent.

Problems

A series RLC circuit has \(R = 100\ \Omega\), \(L = 1\ \mathrm{H}\), and \(C = 1\ \mu\mathrm{F}\). Find the resonant frequency in Hz.
Solution
\(\omega_0 = 1/\sqrt{LC} = 1/\sqrt{1 \times 10^{-6}} = 1000\ \mathrm{rad/s}\). \(f_0 = \omega_0/(2\pi) = 1000/(2\pi) \approx 159.2\ \mathrm{Hz}\).
For the circuit in Problem 1, find the impedance magnitude at \(f = 100\ \mathrm{Hz}\).
Solution
\(\omega = 2\pi(100) \approx 628.3\ \mathrm{rad/s}\). \(X_L = \omega L = 628.3\ \Omega\). \(X_C = 1/(\omega C) = 1/(628.3 \times 10^{-6}) \approx 1591.5\ \Omega\). \(X = X_L - X_C = 628.3 - 1591.5 = -963.2\ \Omega\). \(|Z| = \sqrt{R^2 + X^2} = \sqrt{100^2 + (-963.2)^2} \approx \sqrt{10000 + 927754} \approx 968.4\ \Omega\).
A series RLC circuit is driven by a \(240\ \mathrm{V}\) (RMS) source at resonance. If \(R = 20\ \Omega\), find the RMS current and the power dissipated.
Solution
At resonance, \(Z = R = 20\ \Omega\). \(I_{\text{rms}} = V_{\text{rms}}/R = 240/20 = 12\ \mathrm{A}\). Power \(P = I_{\text{rms}}^2 R = 12^2 \times 20 = 144 \times 20 = 2880\ \mathrm{W}\).
For a series RLC circuit with \(R = 50\ \Omega\), \(L = 0.2\ \mathrm{H}\), and \(C = 50\ \mu\mathrm{F}\), find the quality factor \(Q\) and the bandwidth \(\Delta f\) in Hz.
Solution
\(\omega_0 = 1/\sqrt{LC} = 1/\sqrt{0.2 \times 50 \times 10^{-6}} = 1/\sqrt{10^{-5}} \approx 316.2\ \mathrm{rad/s}\). \(Q = \omega_0 L / R = 316.2 \times 0.2 / 50 = 63.24 / 50 \approx 1.26\). \(\Delta\omega = R/L = 50/0.2 = 250\ \mathrm{rad/s}\). \(\Delta f = \Delta\omega/(2\pi) = 250/(2\pi) \approx 39.8\ \mathrm{Hz}\).
A series RLC circuit has \(R = 10\ \Omega\), \(L = 0.1\ \mathrm{H}\), and \(C = 100\ \mu\mathrm{F}\). The source voltage is \(V(t) = 10 \cos(314 t)\ \mathrm{V}\). Find the expression for the current \(I(t)\).
Solution
\(\omega = 314\ \mathrm{rad/s}\). \(X_L = 314 \times 0.1 = 31.4\ \Omega\). \(X_C = 1/(314 \times 100 \times 10^{-6}) \approx 31.85\ \Omega\). \(X = 31.4 - 31.85 = -0.45\ \Omega\). \(|Z| = \sqrt{10^2 + (-0.45)^2} \approx 10.01\ \Omega\). \(\phi = \arctan(-0.45/10) \approx -2.6^\circ\). \(I_0 = 10/10.01 \approx 1.00\ \mathrm{A}\). Since \(\phi\) is negative, current leads voltage. \(I(t) = 1.0 \cos(314 t + 2.6^\circ)\ \mathrm{A}\).