An AC Circuit with Just an Inductor

In a circuit with only an inductor and an AC power source, the inductor voltage has to match the source voltage at all times. This comes directly from Kirchoff's loop rule:
Vo sin(wt) – L
dI
dt
= 0
So, L
dI
dt
= Vo sin(wt)

When is the current at its peak magnitude in this circuit? Once again you might find it helpful to ask the equivalent question for an object oscillating on a spring. The current peaks ____

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














The current peaks when the voltage passes through zero. To achieve a zero voltage the slope of the current graph must be zero - that's at a maximum or minimum of the current vs. time graph. The equivalent question for an oscillating object is, when does the velocity peak, when the force is maximum or when the force passes through zero?

What affect would increasing the inductance have on the peak current in the circuit?

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














Increasing L means that the inductor is more effective at opposing any changes in current. This decreases the current in the circuit.

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 source is changing more rapidly. The inductor opposes change, so if there is more change there is more opposition from the inductor. The current goes down.

For a mathematical analysis, start with:
dI
dt
=
Vo
L
sin(wt)

Integrate to find the current as a function of time:
I(t) = –
Vo
wL
cos(wt)

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

We often write the peak current in the form:
Io =
Vo
wL
=
Vo
XL

where XL = wL is the effective resistance of the inductor and is known as the inductive reactance.

The peak current is inversely 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 across the inductor. The higher the frequency, the smaller the time interval over which the voltage changes, and the smaller DI has to be.