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AC Circuits

# Capacitance in AC Circuits

## The Pure Capacitance Circuit

A change in voltage across a capacitor results in a current that is proportional to both the rate of voltage change and the capacitance. The relations are given by the equation

A sinusoidal voltage is applied to a capacitor in the circuit of the figure
above. The current is plotted versus time with reference to voltage *v*
in the figure below.

The plot of voltage in the figure above increases from *t*_{0} to
*t*_{1}. The voltage is increasing at a decreasing rate, and at
*t*_{1} the rate of change of voltage is zero. At time
*t*_{1}, the current must then be zero. From time
*t*_{1} to *t*_{2}, the voltage is decreasing, and
at *t*_{2} the voltage is changing at a maximum rate. The current
is negative from *t*_{1} to *t*_{3} and maximum
negative at *t*_{2}. At *t*_{3}, the rate of change of
voltage is instantaneously zero, and therefore the current is zero.
From *t*_{3} to *t*_{4}, the voltage is increasing
at an increasing rate, and maximum instantaneous rate of change occurs at
*t*_{4}. Therefore *i*
is maximum at *t*_{4}.

From the waveforms of figure above, it is seen that the maximum positive current occurs 90 degrees ahead of the maximum positive voltage. The current is said to be leading the voltage by 90° in a purely capacitive circuit. This phase relationship is derived mathematically by applying the calculus.

Using the equation

and

By differentiation, *dv*/*dt* is found.

Therefore,

By the general form of a periodic function,

and

Since the ratio of volts to amperes is defined as opposition to current in ohms,
the quantity 1/ωC is in ohms. The quantity 1/ωC is called
*capacitive reactance* and is symbolized *X*_{C}.

*X*_{C} can be shown to be the ratio of effective values of
current and voltage in the same manner as was shown for *X*_{L}
in the previous section.

The reciprocal of capacitive reactance is called *capacitive susceptance*
and is symbolized *B*_{C}. The unit of capacitive susceptance
is the mho (or siemens S) when the frequency is in Hz and the capacitance is
in farads.

**Example 1**: A voltage having an effective value of 220 volts and a
frequency of 20 kHz is applied to an 0.08-μF capacitor. (1) Calculate
the effective value of the current. (2) Calculate the maximum value of the
current. (3) Write the periodic functions representing current and voltage.

Solution:

1.

2.

3. If the voltage is taken as a reference,

The current leads the voltage by 90° in a capacitive circuit.

These periodic functions can also be written with reference to the current.

With reference to the current, the voltage lags by 90°.

**Example 2**: When 250 volts is applied to a 0.05μF-capacitor,
a current of 0.6 A is measured. Find the frequency.

Solution: The magnitude of *X*_{C} can be found.

The frequency can now be calculated from

The power relation in a capacitive circuit can be analyzed with the
equation *P*=*VI* on the basis of instantaneous values. The product
of *v* and *i* of the figure above is plotted in the figure below.
This is the same as was done in the preceding section for instantaneous
power in a pure inductive circuit. The net power in a purely capacitive
circuit is zero, as it was in the purely inductive circuit.

This relationship is justified by considering that from *t*_{0}
to *t*_{1} the voltage is increasing, charge is being stored,
and energy is being stored by the capacitor. From *t*_{1} to
*t*_{2}, the voltage is decreasing, charge is
returned by the capacitor, and energy is returned to the source.

The net power is again given by the equation *P* = *VI* cos *θ*