Direct current (DC) circuits involve current flowing in one direction. In alternating current (AC) circuits the voltage oscillates in a sine wave pattern:

V(t) = V_o sin(ωt)

In a household circuit, the frequency is 60 Hz. The angular frequency is related to the frequency, f, by:

ω = 2πf

V_o represents the maximum voltage, which in a household circuit in North America is about 170 volts. We talk of a household voltage of 120 volts, though; this number is a kind of average value of the voltage. The particular averaging method used is called root mean square (square the voltage to make everything positive, find the average, take the square root), or rms. Voltages and currents for AC circuits are generally expressed as rms values. For a sine wave, the relationship between the peak and the rms average is:

rms value = 0.707 peak value

RMS averaging

The average of the numbers -1, 1, 3, and 5 is 2, while their rms average is 3. Root mean square averaging turns everything positive, and weights larger numbers more than smaller numbers.

To determine the rms average of sin(θ) over one period, first square it and then find the mean. This can be done by integrating:

Average value of sin²(θ) = [∫ sin²(θ) dθ ]/π

where the limits on the integral are from 0 to π

Use the trig. identity
sin²(θ) = 1/2[1 - cos(2θ)]

The integral of cos(2θ) over the interval 0 to π is zero, so we get:

Average value of sin²(θ) = 1/2π ∫ dθ = 1/2

Finally, take the square root:

rms value of a sine wave is 1/2^1/2 = 0.707

The relationship $Delta;V = IR applies for resistors in an AC circuit, so the current is:

I(t) = (V_o/R) sin(ωt)

The peak current is the peak voltage over the resistance, and does not depend on frequency. In addition, the current is in phase with the voltage.

Consider now a circuit which has only a capacitor and an AC power source (such as a wall outlet). A capacitor is a device for storing charging. It turns out that there is a 90^o phase difference between the current and voltage, with the current reaching its peak 90^o (1/4 cycle) before the voltage reaches its peak. Put another way, the current leads the voltage by 90^o in a purely capacitive circuit.

To understand why this is, simply apply Kirchoff's loop rule:

V_o sin(ωt) - Q/C = 0

Q = CV_o sin(ωt)

I(t) = dQ/dt = ωCV_o cos(ωt)

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:

I_o = ωCV_o = V_o / X_L

where X_C = 1/ωC is the effective resistance of the capacitor and is known as the capacitive reactance.

Note that the peak current is inversely 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.

In a circuit with only an inductor and an AC power source, there is also a 90^o phase difference between the current and voltage. This time the current lags the voltage by 90^o in a purely inductive circuit.

To understand why this is, apply Kirchoff's loop rule:

V_o sin(ωt) - L dI/dt = 0

dI/dt = V_o/L sin(ωt)

Integrate to find the current as a function of time:

I(t) = -(V_o/ωL) cos(ωt)

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

We often write the peak current in the form:

I_o = V_o/ωL= V_o / X_L

where X_L = ωL is the effective resistance of the inductor and is known as the inductive reactance.

The peak current is proportional to both the angular frequency and the inductance. We can understand this by realizing that the inductor voltage has to match the source voltage at all times. The larger the inductance, the smaller the rate of change of current necessary to achieve a particular voltage on the plates. The higher the frequency, the smaller the time interval over which the voltage changes, and the smaller $Delta;I has to be.