Physics · Ch 4 — Electromagnetic Induction and Alternating Current

AC Circuit Containing Only a Capacitor

4.7.7

AC Circuit Containing Only a Capacitor

Consider a circuit containing a capacitor of capacitance C connected directly across an alternating voltage source $v=V_m\sin\omega t \qquad (4.43)$ . Letting q be the instantaneous charge on the capacitor, the emf across it at that instant is $q/C$ ; Kirchhoff's loop rule gives $v-q/C=0$ , so $q=CV_m\sin\omega t$ . By definition, the current is $i=dq/dt=CV_m\,d(\sin\omega t)/dt = CV_m\omega\cos\omega t$ , which can be rewritten using $\cos\theta=\sin(\theta+\pi/2)$ as

i = \dfrac{V_m}{1/\omega C}\sin\left(\omega t + \dfrac{\pi}{2}\right) = I_m\sin\left(\omega t+\dfrac{\pi}{2}\right) \qquad (4.44)

where $I_m=V_m/(1/\omega C) = V_m\omega C$ is the peak current. Comparing with the applied voltage, the current in a purely capacitive circuit LEADS the voltage by exactly $90^{\circ}$ -- the mirror image of the inductive circuit's lag -- confirmed on both the phasor diagram (current phasor $90^{\circ}$ ahead of the voltage phasor) and the wave diagram (current's peaks and zero-crossings each occurring a quarter cycle earlier than voltage's). This lead relationship is remembered by the mnemonic ICE -- current leads EMF (voltage) in a Capacitor.

Capacitive reactance $X_C$ . Comparing $I_m=V_m/(1/\omega C)$ with $I_m=V_m/R$ shows that $1/\omega C$ plays the role of resistance for a capacitor, called the capacitive reactance,

$X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C} \qquad (4.45)$ …

Figure 4.44AC circuit with a capacitor

What this figure shows. A capacitor C 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 inductor present. Kirchhoff's loop rule applied here, $v-q/C=0$ , gives the instantaneous charge $q=CV_m\sin\omega t$ , and differentiating using $i=dq/dt$ produces the circuit's central result -- that the current in a purely capacitive AC circuit LEADS the applied voltage by exactly $90 …

Figure 4.45Phasor and wave diagram for an AC circuit with C -- current leads 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 $90^{\circ}$ AHEAD of 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 EARLIER than the voltage curve's. The figure is the mirror image of Figure 4.43 for the inductive case: $i=I_m\sin(\omega t+\pi/2)$ , current running a quarter cycle ahead of voltage, summarised by the ICE mnemonic (current leads EMF, i.e. voltage, …

Misc Example 4.21Capacitive reactance, RMS current and equations for a capacitor on the mains

Worked out. A capacitor of capacitance $\dfrac{10^{-4}}{2\pi}$ F is connected across a 220 V, 50 Hz AC mains, and the capacitive reactance, RMS current, and the equations of voltage and current are required. The capacitive reactance is $X_C = 1/(2\pi fC) = 1/\left(2\pi\times50\times\dfrac{10^{-4}}{2\pi}\right) = 1/(50\times10^{-4}) = 100\ \Omega$ . The RMS current is $I_{RMS}=V_{RMS}/X_C = 220/100 = 2.2$ A. The peak values are $V_m=220\sqrt2\approx311$ V and $I_m=2.2\sqrt2\approx3.1$ A, so the two equations are $v=311\sin(314t)$ V and, since current leads voltage by $90^{\circ}$ in a pure capacitor, $i=3.1\sin(314t+\pi/2)$ A. This worked example strings together the capacitive reactance formula, the RMS-to-peak conversion, and the fixed $90^{\circ}$ phase-lead r …

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