17.8 AC Through a Capacitor

17.8 AC Through a Capacitor

A capacitor does not allow direct current to pass through it because of the presence of an insulating medium between its plates. But alternating current can pass through a capacitor. In electric circuits, a capacitor is a reactive component. Unlike a resistor, a capacitor behaves differently in AC and DC circuits. It is because a capacitor can store energy in the form of electric field, whereas a resistor cannot store electrical energy in any form. Consider an alternating voltage source $V$ is applied to a capacitor of capacitance $C$ as shown in Fig. 17.8 (a). When an alternating voltage is applied across the plates of a capacitor, it is charged in one direction and then in the other as the voltage reverses. The result is that electrons move to and fro around the circuit, connecting the plates, thus, constituting alternating current. The basic relationship between charge $q$ and the voltage $V$ across the plates: $q = CV$, holds good at every instant. Let $V$ be the applied alternating voltage given by

(Fig. 17.8(a): Showing capacitor connected with ac voltage source)

The change on the capacitor at any instant is give by $q = CV = CV_{o} \sin\omega t$ Thus $I = \dfrac{\Delta q}{\Delta t} = \dfrac{\Delta}{\Delta t} (CV_{o} \sin\omega t)$ $I = CV_{o} \dfrac{\Delta}{\Delta t} (\sin\omega t)$ As $\dfrac{\Delta}{\Delta t} \sin\omega t = \omega \cos\omega t$ , thus $I = CV_{o} \omega \cos\omega t$ As $\cos(\omega t) = \sin\left(\omega t + \dfrac{\pi}{2}\right)$ , so $I = \omega CV_{o} \sin\left(\omega t + \dfrac{\pi}{2}\right)$ $I = \omega CV_{o} \sin\left(\omega t + \dfrac{\pi}{2}\right)$ Here $\omega C V_{o} = I_{o}$ , thus

(Fig. 17.8 (b): Graphical representation of current and voltage for capacitor.)

(Fig. 17.8©: Vector representation of current and voltage for a capacitor.)

Equations 17.18 and 17.19 show that inductor current $I$ leads the voltage by $90^{\circ}$ or $\pi/2$ rad or voltage lags behind the current $I$ by $90^{\circ}$ or $\pi/2$ rad as shown in Figs. 17.8(b) and .17.8(c). Resistance offered by capacitor is known as capacitive reactance denoted by $X_{C}$ and is given by $X_{C} = \dfrac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}}$ where $V_{\mathrm{rms}}$ is the rms value of alternating voltage in the inductor and $I_{\mathrm{rms}}$ is the rms current passes through it. As $V_{\mathrm{rms}} = 0.707 V_{o}$ and $I_{\mathrm{rms}} = 0.707 I_{o}$, so $X_{C} = \dfrac{0.707 V_{o}}{0.707 I_{o}} = \dfrac{V_{o}}{I_{o}}$ As $I_{o} = \omega CV_{o}$ Hence $X_{C} = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$ where $\omega = 2\pi f$ The unit of capacitive reactance $X_{C}$ is ohm and is given as: $X_{C} = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$ Which shows that for low frequencies, capacitor will have a large reactance $X_{C}$ and current $I$ will be small whereas at high frequencies reactance $X_{C}$ will be small and current $I$ through the same capacitor will be large.

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Father of ALTERNATING CURRENTS? Nikola Tesla was born in 1856 in Austria-Hungary and emigrated to the U.S.A. in 1884 as a physicist. He pioneered the generation, transmission, and use of alternating current which can be transmitted over much greater distances than direct current.