Physics · Ch 4 — Electromagnetic Induction and Alternating Current

AC Circuit Containing Only an Inductor

4.7.6

AC Circuit Containing Only an Inductor

Consider a circuit containing a pure inductor of inductance L connected directly across an alternating voltage source $v=V_m\sin\omega t \qquad (4.40)$ . The current flowing through the inductor induces a self-induced back emf $\varepsilon=-L\,di/dt$ ; applying Kirchhoff's loop rule to this purely inductive circuit, $v+\varepsilon=0$ , gives $V_m\sin\omega t = L\,di/dt$ , i.e. $di=(V_m/L)\sin\omega t\,dt$ . Integrating both sides (and taking the time-independent integration constant as zero, since the applied voltage has no steady/DC component), $i=(V_m/L)\int\sin\omega t\,dt = -(V_m/\omega L)\cos\omega t = (V_m/\omega L)\sin(\omega t-\pi/2)$ (using $-\cos\theta=\sin(\theta-\pi/2)$ ), giving

i = I_m\sin\left(\omega t - \dfrac{\pi}{2}\right) \qquad (4.41)

where $I_m=V_m/(\omega L)$ is the peak current. Comparing with the applied voltage, the current in a purely inductive circuit LAGS the voltage by exactly $90^{\circ}$ ( $\pi/2$ radians) -- confirmed on both the phasor diagram (current phasor $90^{\circ}$ behind the voltage phasor) and the wave diagram (current's peaks and zero-crossings each occurring a quarter cycle later than voltage's). This lag relationship is commonly remembered by the mnemonic ELI -- EMF (voltage) Leads current in an Inductor.

Inductive reactance $X_L$ . Comparing $I_m=V_m/(\omega L)$ with the resistive-circuit relation $I_m=V_m/R$ shows that the quantity $\omega L$ plays exactly the role R plays in a resistive circuit -- the effective opposition (measured in ohms) that an inductor offers to alternating current, called the inductive reactance,

$X_L = \omega L = 2\pi f L \qquad (4.42)$ …

Figure 4.42AC circuit with an inductor

What this figure shows. A pure inductor L is connected directly across an alternating source labelled $v=V_m\sin\omega t$ , drawn as a simple single-loop circuit with no resistor or capacitor present. Kirchhoff's loop rule applied to this circuit, $v+\varepsilon=0$ with the inductor's back emf $\varepsilon=-L\,di/dt$ , sets up the differential equation $V_m\sin\omega t = L\,di/dt$ that, once integrated, produces the circuit's central result -- that the current in a purely inductive AC circuit lags $90^{\circ}$ behind th …

Figure 4.43Phasor and wave diagram for an AC circuit with L -- current lags voltage by 90°

What this figure shows. Voltage phasor $\vec{OA}$ (length $V_m$ ) is drawn along the reference axis, with current phasor $\vec{OB}$ (length $I_m$ ) positioned $90^{\circ}$ BEHIND it in the anticlockwise rotation sense, alongside matching sine-wave graphs of v and i showing the current curve's peaks and zero-crossings each occurring a quarter cycle LATER than the voltage curve's. The figure is the visual proof of the inductive circuit's phase result: $i=I_m\sin(\omega t-\pi/2)$ , current trailing voltage by exactly a quarter cycle, summarised by the ELI mnemonic (EMF Leads current i …

Misc Example 4.20Voltage across a 400 mH coil carrying a given RMS current

Worked out. A 400 mH coil of negligible resistance carries an effective (RMS) current of 6 mA at frequency 1000 Hz, and the voltage across the coil is required. The inductive reactance is $X_L=\omega L = 2\pi f L = 2\pi(1000)(0.4)\approx2512\ \Omega$ . Since RMS voltage and RMS current obey the same reactance-based relation as peak values, $V_{RMS}=I_{RMS}X_L=(6\times10^{-3})(2512)\approx15.07$ V. The example is a direct, one-step application of the inductive-reactance formula, working entirely with RMS (rather than peak) values throughout, exactly as an ammeter or voltmeter would actuall …

Misc Example 4.25Writing the voltage equation from a given inductive current

Worked out. The current in an inductive circuit is given by $i=0.3\sin(200t-40^{\circ})$ A, and the voltage equation across it is required, given inductance 40 mH. The inductive reactance is $X_L=\omega L = 200\times40\times10^{-3}=8\ \Omega$ , so the peak voltage is $V_m=I_mX_L=0.3\times8=2.4$ V. Because voltage LEADS current by $90^{\circ}$ in a purely inductive circuit, the voltage's phase angle is the current's phase angle PLUS $90^{\circ}$ : $v=V_m\sin(\omega t+\phi_i+90^{\circ})=2.4\sin(200t-40^{\circ}+90^{\circ})=2.4\sin(200t+50^{\circ})$ V. The example reverses the usual direction of reasoning -- starting from a given current equation and working out the corresponding voltage equation -- by combining the reactance formula with the …

Start your 14-day free trial to unlock the full solution →

← Previous · 4.7.5AC Circuit Containing a Pure Resistor Next · 4.7.7 →AC Circuit Containing Only a Capacitor