AC Voltage Applied to an Inductor

What this figure shows. A simple series circuit diagram showing a pure inductor (coil symbol, L) connected across an AC source, with a switch/key k in the loop, and the instantaneous emf e and instantaneous current i labelled -- the purely inductive analogue of Fig. 13.3, with the inductor assumed to have negligible ohmic resistance so that the only opposition to current comes from the self-induced back-emf $e'=-L\,di/dt$ (Lenz's law) that opposes the applied emf as the key is closed …

What this figure shows. Two sinusoidal curves on the same axes (e and i vs $\omega t$, one full cycle), both of the same general sine shape but HORIZONTALLY OFFSET from each other by a quarter cycle ($\pi/2$ radian): the e curve peaks at $\omega t=\pi/2$ as usual, while the i curve (current) reaches ITS peak later, at $\omega t=\pi$ (i.e. i is delayed/lags behind e by $\pi/2$) -- so wherever e crosses zero going upward, i is still at its most negative, and i only reaches zero a quarter-cycle after e does. The curves visually establish the $90^\circ$ lag of current behind emf that is the defining elec …

What this figure shows. A phasor diagram with fixed horizontal reference axis OX, showing the peak-emf phasor $e_0$ drawn at angle $\omega t$ anticlockwise from OX (as in the general phasor convention), and the peak-current phasor $i_0$ drawn turned CLOCKWISE by $90^\circ$ from the $e_0$ phasor's direction (i.e. lagging behind it), so the angle between the two phasor arrows is exactly $90^\circ$ with $i_0$ trailing $e_0$ -- visually encoding 'current lags voltage by $90^\circ$ in a pure inductor', the mirror image of the capacitor's phasor diagram (Fig. 13.11) wher …

Worked out. An inductor of inductance $L=200$ mH $=0.2$ H is connected to an AC source of peak emf $e_0=210$ V and frequency $f=50$ Hz. The peak current is found via $i_0=e_0/X_L=e_0/(2\pi fL)=210/(2\times3.142\times50\times0.2)\approx3.342$ A. For the second part, since current LAGS emf by $\pi/2$ in a pure inductor, the instant at which current reaches its own peak value is exactly the instant at which the emf phasor has already rotated a further $\pi/2$ past where it was when current started rising from zero -- concretely, at the moment $i=i_0$, the emf phasor is at the position where $\sin(\cdot)=0$, so the instantaneous voltage of the source at that moment is zero. This example fixes the practical consequence of the $90^\circ$ lag: peak current and peak voltage NEVER occur at the same …