AC Voltage Applied to a Capacitor

What this figure shows. A simple series circuit diagram showing a capacitor (two parallel-plate symbol, C) connected across an AC source, with the instantaneous emf e and instantaneous current i labelled on the connecting wires -- the purely capacitive analogue of Figs. 13.3 and 13.6, illustrating that as the source current reverses direction every half cycle, the capacitor plates are alternately charged and then discharged/recharged with opposite polarity, so a continuous alternating current flows in the external circuit even though no charge actu …

What this figure shows. Two sinusoidal curves on the same axes (e and i vs $\omega t$, one full cycle), horizontally offset from each other by a quarter cycle in the OPPOSITE sense to the inductor case: the current curve i reaches ITS peak EARLIER than the emf curve e, at $\omega t=0$ (or equivalently the i-curve is the e-curve shifted to the LEFT by $\pi/2$) -- so i crosses zero and reaches its peak a quarter-cycle BEFORE e does at each corresponding point, visually establishing the $90^\circ$ LEAD of current ahead of emf that is the defining electrical signature of a pure capacitor, the mirror …

What this figure shows. A phasor diagram with fixed horizontal reference axis OX, showing the peak-emf phasor $e_0$ at angle $\omega t$ anticlockwise from OX, and the peak-current phasor $i_0$ drawn turned a FURTHER $90^\circ$ ANTICLOCKWISE ahead of the $e_0$ phasor's direction (i.e. leading it), so the current phasor arrow sits $90^\circ$ ahead of the emf phasor arrow, both rotating together anticlockwise at the same rate $\omega$ while maintaining this fixed $90^\circ$ separation -- the mirror image of the inductor's phasor diagram (Fig. 13.8), wher …

Resistance | Reactance

Equally effective for AC and DC | Current is affected (reduced) but energy is not consumed (heat is not generated); the energy consumption by a real coil is due to its resistive component only

Its value is independent of the frequency of the AC | Inductive reactance ($X_L=2\pi fL$) is directly proportional, and capacitive reactance ($X_C=\frac{1}{2\pi fC}$) is inversely proportional, to the frequency of the AC …

Worked out. A capacitor of $C=2\,\mu F=2\times10^{-6}$ F is connected to an AC source of emf $e=250\sin(100\pi t)$. Since current leads emf by $\pi/2$ in a pure capacitor, the instantaneous current has the form $i=i_0\sin(\omega t+\pi/2)$ with peak $i_0=\omega C\,e_0=(100\pi)(2\times10^{-6})(250)=3.142\times2\times10^{-4}\times250\approx0.1571$ A, giving $i=0.1571\sin(100\pi t+\pi/2)$ A. An AC ammeter reads the rms value: $i_{rms}=0.707\,i_0=0.707\times0.1571\approx0.111$ A. This example demonstrates reading the peak current directly off $i_0=\omega C\,e_0$ (equivalently $e_0/X_C$) and then converting to the rms value that a real …