Phase Relation in a Pure Capacitor

The relation between voltage and current in the case of a capacitor is given by:

$$i(t) = C\frac{dv(t)}{dt}$$

where:

• $C$ = capacitance
• $v(t)$ = voltage
• $i(t)$ = current

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Sinusoidal Voltage Applied to a Capacitor

Consider the voltage function:

$$v(t) = V_m \sin \omega t$$

Differentiating with respect to time,

$$\frac{dv(t)}{dt} = V_m \omega \cos \omega t$$

Substituting into the current equation,

$$i(t) = CV_m \omega \cos \omega t$$

Using the trigonometric identity,

$$\cos \omega t = \sin\left(\omega t + 90^\circ\right)$$

Therefore,

$$i(t) = \omega C V_m \sin\left(\omega t + 90^\circ\right)$$

Let,

$$I_m = \omega C V_m$$

Hence,

$$i(t) = I_m \sin\left(\omega t + 90^\circ\right)$$

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Phase Relationship

If we draw the waveform for both voltage and current as shown in Fig. 2.12, there is a phase difference between these two waveforms.

In a pure capacitor:

The current leads the voltage by $90^\circ$.

or equivalently,

The voltage lags behind the current by $90^\circ$.

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Impedance of a Pure Capacitor

The impedance is the ratio of voltage to current.

$$Z = \frac{V}{I}$$

For a pure capacitor,

$$Z = \frac{V_m}{I_m} = \frac{1}{\omega C}$$

The capacitive reactance is represented by:

$$X_C = \frac{1}{\omega C}$$

where:

• $\omega = 2\pi f$
• $C$ = capacitance

Thus,

$$X_C = \frac{1}{2\pi fC}$$

The impedance of a pure capacitor is represented in complex form as:

$$Z = -jX_C$$

or

$$Z = -\frac{j}{\omega C}$$

Hence, the impedance value of a pure capacitor is:

$$Z = -\frac{j}{\omega C}$$

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Characteristics of Pure Capacitive Circuit

• Current leads voltage by $90^\circ$.
• Impedance is purely imaginary.
• No real power is consumed.
• Energy is temporarily stored in the electric field.
• Average power over one complete cycle is zero.

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Power in a Pure Capacitor

Instantaneous power is:

$$p(t) = v(t)i(t)$$

Average power is:

$$P = VI\cos\phi$$

For a pure capacitor,

$$\phi = 90^\circ$$

Since,

$$\cos 90^\circ = 0$$

therefore,

$$P = 0$$

Thus, a pure capacitor consumes no average power.

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Summary

• In a pure capacitor, current leads voltage by $90^\circ$.
• Voltage lags current by $90^\circ$.
• Capacitive reactance is:

$$X_C = \frac{1}{\omega C}$$

• The impedance of a capacitor is:

$$Z = -\frac{j}{\omega C}$$

• A pure capacitor stores energy in its electric field.
• The average power consumed by a pure capacitor is zero.