Consider a circuit that has only a capacitor and an AC power source (such as a wall outlet). In this situation the capacitor voltage must match the source voltage at all times.

V_c = V_o sin(wt)

Because the voltage is oscillating the capacitor is continually charging and discharging.

When is the current at its peak magnitude in this circuit?

1. When the voltage is at its peak magnitude
2. When the voltage passes through zero

The current peaks when the voltage changes most rapidly, when the voltage passes through zero. To change the capacitor voltage quickly the capacitor must be charging, or discharging, quickly - so the current must be high. To keep the voltage at the same level, on the other hand, requires no current at all.

What affect would increasing the capacitance of the capacitor have on the peak current in the circuit?

1. Increasing C would increase the peak current
2. Increasing C would not affect the peak current
3. Increasing C would decrease the peak current

Increasing C means that to achieve a certain potential difference on the capacitor requires more charge. Changing from one voltage to another requires more charge to flow. So, the peak current would increase.

What affect would increasing the frequency of the voltage have on the peak current in the circuit?

1. Increasing f would increase the peak current
2. Increasing f would not affect the peak current
3. Increasing f would decrease the peak current

Increasing f means the capacitor needs to charge and discharge more rapidly, so that requires a larger current.

For a mathematical analysis, apply Kirchoff's loop rule:

Vo sin(wt)

–

Q C

=

0

Q = CV_o sin(wt)

I(t)

=

dQ dt

=

wCVo cos(wt)

The current follows a cosine graph while the voltage follows a sine graph, explaining why the current leads the voltage by 90^o

We often write the peak current in the form:

Io

=

wCVo

=

Vo XC

where

XC

=

1 wC

X_C is the effective resistance of the capacitor and is known as the capacitive reactance.

Note that the peak current is proportional to both the angular frequency and the capacitance. We can understand this by realizing that the capacitor voltage has to match the source voltage at all times. The larger the capacitance of the capacitor, the more charge has to flow to build up a particular voltage on the plates, and the higher the current will be. The higher the frequency of the voltage, the shorter the time available to change the voltage, so the larger the current has to be.