Physics · Ch 7 — Alternating Current

AC Voltage Applied to a Pure Inductor

7.5

AC Voltage Applied to a Pure Inductor

Setting up the circuit. A pure inductor $L$ (assumed to have no resistance of its own) is connected directly across an AC source supplying $v=v_0\sin\omega t$ .

The inductor's defining relation. An inductor's defining property is that the emf induced across it, opposing any change in the current through it, is $L\,di/dt$ (Section 6.7 of the previous sub-topic). Since the source supplies the full voltage $v$ directly across the (resistance-free) inductor, this means

v = L\frac{di}{dt} \quad\Longrightarrow\quad v_0\sin\omega t = L\frac{di}{dt}

Solving for the current. Rearranging and integrating with respect to time,

di = \frac{v_0}{L}\sin\omega t\;dt \quad\Longrightarrow\quad i = -\frac{v_0}{\omega L}\cos\omega t = \frac{v_0}{\omega L}\sin\!\left(\omega t-\frac{\pi}{2}\right)

using the identity $-\cos\theta=\sin(\theta-\pi/2)$ (the constant of integration is dropped since the current itself is purely alternating, with zero average). Writing $i_0=v_0/(\omega L)$ , this is

i = i_0\sin\!\left(\omega t-\frac{\pi}{2}\right)

Reading off the phase. Comparing this with the applied voltage $v=v_0\sin\omega t$ shows the current lags the voltage by exactly $\pi/2$ radians ( $90^\circ$ ) -- the current reaches its own peak a full quarter-cycle AFTER the voltage reaches its peak. Physically, this delay is the inductor actively resisting any change in current: it takes the induced back emf, and hence a full quarter cycle, for the current to catch up with what the voltage is already doing. …

Figure 1Phasor diagram and waveforms for a pure inductive AC circuit

What this figure shows. The figure has two panels. The LEFT panel is a phasor diagram with two arrows from a common origin: the voltage phasor $V_L$ drawn pointing, say, straight up (at some reference angle), and the current phasor $I_L$ drawn $90^\circ$ CLOCKWISE (i.e. BEHIND/lagging) from $V_L$ , with a small curved arc drawn between the two arrows and labelled $\pi/2$ to mark this quarter-turn phase gap, and a small curved arrow showing both phasors rotating anticlockwise together at speed $\omega$ while always keeping this same $90^\circ$ gap between them. The RIGHT panel is a waveform graph with $\omega t$ on the horizontal axis showing two sine curves, $v$ and $i$ , of the same frequency but with the CURRENT curve shifted to the RIGHT of the voltage curve by a quarter cycle (so the current reaches its own peak a quarter-cycle AFTE …

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