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AC Voltage Applied To A Capacitor | AC Circuit Containing Capacitor Only

AC Voltage Applied To A Capacitor | AC Circuit Containing Capacitor Only :- Suppose a pure capacitor is connected to an alternating voltage source, as shown in figure (a) below :

Let the alternating voltage is represented by the following equation :

$\displaystyle E={{E}_{0}}sin\text{ }\omega t$ …..(1)

When an alternating emf is applied across the terminals of a capacitor, an alternating current flows in the circuit. The two plates of the capacitor become positively and negatively charged alternately and continuously, and the amount of charge on the plates varies sinusoidally with time. Accordingly, the electric field between the plates also varies sinusoidally with time.

Suppose at some time t the charge on the plates of the capacitor is q. Therefore, the potential difference between the plates of the capacitor is,

$\displaystyle V=\frac{q}{C}=E={{E}_{0}}sin\text{ }\omega t$

$\displaystyle \Rightarrow q=C{{E}_{0}}sin\text{ }\omega t$

If at any time t the value of current in the circuit is I, then,

$\displaystyle I=\frac{dq}{dt}=\frac{d}{dt}\left( C{{E}_{0}}sin\text{ }\omega t \right)$

$\displaystyle \Rightarrow I=C{{E}_{0}}\left( cos\text{ }\omega t \right)\omega$

$\displaystyle \Rightarrow I=\frac{{{E}_{0}}}{\left( {}^{1}\!\!\diagup\!\!{}_{\omega C}\; \right)}cos\text{ }\omega t$

$\displaystyle \therefore I=\frac{{{E}_{0}}}{\left( {}^{1}\!\!\diagup\!\!{}_{\omega C}\; \right)}sin\left( \omega t+\frac{\pi }{2} \right)$ …..(2)

[Here we have taken $\displaystyle cos\text{ }\omega t=sin\left( \omega t+\frac{\pi }{2} \right)$ and not used $\displaystyle cos\text{ }\omega t=sin\left( \frac{\pi }{2}\,-\omega t \right)$ . The reason is that we need to compare the current in equation (2) with the alternating voltage E in equation (1). Since the expression for the alternating voltage E in equation (1) contains +ωt and not -ωt, we have therefore used $\displaystyle cos\text{ }\omega t=sin\left( \omega t+\frac{\pi }{2} \right)$ .]

The maximum value of the current,

$\displaystyle {{I}_{0}}=\frac{{{E}_{0}}}{\left( {}^{1}\!\!\diagup\!\!{}_{\omega C}\; \right)}$ …..(3)

Hence, the electric current flowing in the circuit,

$\displaystyle I={{I}_{0}}sin\left( \omega t+\frac{\pi }{2} \right)$ …..(4)

From equations (1) and (4),

Phase of the applied alternating emf = ωt

Phase of alternating electric current = ωt + (π/2)

Thus, in a purely capacitive circuit containing only a capacitor , the alternating current leads the alternating electromotive force (emf) by a phase angle of (π/2). This fact is illustrated in the phasor diagram in figure (b) and in the waveform diagram in figure (c).

Capacitive Reactance – X_C

By comparing equation (3) with Ohm’s law, we find that the effective resistance offered by the capacitor C in the path of electric current is (1/ωC). Therefore, 1/ωC is called the capacitive reactance. Hence, the capacitive reactance is :

$\displaystyle {{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2\pi \nu C}$ …..(5)

Where is the frequency of the alternating voltage source. For a given capacitor,

$\displaystyle {{X}_{C}}\propto \frac{1}{\nu }$ …..(6)

Thus, the capacitive reactance is inversely proportional to the frequency; therefore, the graph between X_C and ν is obtained as follows :

For direct current (DC), the frequency ν = 0. Hence, $\displaystyle {{X}_{C}}=\infty$ , meaning that a pure capacitor offers infinite resistance in the path of direct current. In other words, a pure capacitor cannot conduct direct current.

Unit of Capacitive Reactance

$\displaystyle {{X}_{C}}=\frac{1}{\omega C}\Rightarrow \sec .\left( \frac{1}{farad} \right)=\sec .\left( \frac{Volt}{Coulomb} \right)=\frac{Volt\sec .}{Amp.\sec .}=\frac{Volt}{Amp.}=ohm$

Therefore, the SI unit of inductive reactance is ohm (Ω).

Capacitive Susceptance (S_C)

The reciprocal of capacitive reactance is called capacitive susceptance or capacitive conductance.

$\displaystyle {{S}_{\text{C}}}=\frac{1}{{{X}_{C}}}=\omega C$ …..(7)

The SI unit of S_C is Ω^-1 or mho.

Average Power

(AC Voltage Applied To A Capacitor | AC Circuit Containing Capacitor Only)

The instantaneous power of a circuit containing a pure capacitor,

$\displaystyle P=E\times I=\left( {{E}_{0}}sin\text{ }\omega t \right)\times \left[ {{I}_{0}}sin\left( \omega t+\frac{\pi }{2} \right) \right]$

$\displaystyle \Rightarrow P={{E}_{0}}{{I}_{0}}\left( sin\text{ }\omega t \right)\times \left[ sin\left( \omega t+\frac{\pi }{2} \right) \right]={{E}_{0}}{{I}_{0}}\left( sin\text{ }\omega t \right)\left( cos\text{ }\omega t \right)$

$\displaystyle \Rightarrow P={{E}_{0}}{{I}_{0}}\left( sin\text{ }\omega t \right)\left( cos\text{ }\omega t \right)$

$\displaystyle \Rightarrow P=\frac{{{E}_{0}}{{I}_{0}}}{2}\left( 2\text{ }sin\omega t\text{ }cos\omega t \right)$

$\displaystyle \Rightarrow P=\frac{{{E}_{0}}{{I}_{0}}}{2}sin2\omega t$

Now, over a complete cycle, the value of (sin 2ωt) is zero. Therefore, the average power of the circuit over one complete cycle is :

$\displaystyle {{P}_{avg.}}=\langle \frac{{{E}_{0}}{{I}_{0}}}{2}sin2\omega t\rangle =\frac{{{E}_{0}}{{I}_{0}}}{2}\langle sin2\omega t\rangle =\frac{{{E}_{0}}{{I}_{0}}}{2}\times 0=0$

Thus, in a complete cycle, the average power supplied to a pure capacitor is zero.

Example 1.

(NCERT Example 7.3)

A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced?

Solution :

When a capacitor is connected to a DC source, the capacitor gets charged, and after it is fully charged, no current flows in the circuit, and the lamp does not light up. In this situation, reducing the capacitance will not cause any change.

When a capacitor is connected in a circuit with an AC source, the capacitive reactance of the capacitor ( $\displaystyle {{X}_{C}}=\frac{1}{\omega C}$ ) comes into play, allowing current to flow through the circuit. As a result, the lamp will glow. If the capacitance is decreased, the capacitive reactance increases, and the lamp will glow with less brightness compared to before.

Example 2.

(NCERT Example 7.4)

A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current ?

Solution :

Capacitive reactance,

$\displaystyle {{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2\pi \nu C}=\frac{1}{2\pi \times 50\times 15\times {{10}^{-6}}}=212.3\Omega$