Suppose I have the following circuit:

I close the switch. In steady-state the current goes to I = V/R. How long does it take for I to reach this value? For example, does it do it instantaneously? No, because of Faraday's law, we see that the following occurs:

• The switch is closed and I starts to flow.
• The I produces a B-field which threads through the circuit. Since I is increasing, this means that an emf is set up which drives a current which opposes the increase in the magnetic flux.
• This corresponds to the addition of "battery" to the circuit as

  

• This induced emf slows down the gorwth of the induced current and is known as
  back-emf

So, what is the strength of the self-inductance?

Faraday's Law is
emf = - N delta(Magnetic Flux)/delta(time)

For the circuit shown above (and in general), the flux threading the loop depends on B which is proportional to the size of the current, i.e.,

B proportional to I

So, we then have that the induced emf is going to be given by

emf = - L delta(I)/delta(t)

where we have absorbed the "physical" terms (areas, material properties, ...) into a new quantity for the circuit, L. L is called the self-inductance. L is a measure of the resistance to change. It has units of Henries = 1 V-s/A.

The determination of L is usually not simple and must be found in the laboratory. However, let's look at a solenoid. The B-field of a long solenoid is
B = mu n I

where mu is the permeability of the medium, n is the number of turns of the solenoid per meter and I is the current running through the solenoid. The magnetic flux through one loop of the the solenoid is then
Flux = Phi = mu x n x I x Area = (N/length) x mu x I x Area

Now suppose that we start from the situation where I = Phi = 0 at the start. Then, we simply have

delta(Phi) = Phi(t) - 0 = Phi(t) and delta(i) = I(t) - 0 = I(t)

and so, we have that

emf = - N x delta(Phi)/delta(t) = -N x Phi/t = L x I/t ===> L = -N x Phi/I

The inductance is then

L = - (N^2/length) x mu x Area x I / I = - (N^2/length) x mu x Area

Thus, as claimed, the inductance, L, only depends upon the physical properties of the solenoid.

R-L Circuit



Given the above circuit, close the switch at time t = 0 s. What happens?

• A current I starts to flow. How big is the current, well, if L is constant then,
  V = I x R + L x delta(I)/delta(t) ===> I = V/R - (L/R) x dI/dt

  ===> I = (V/R) (1 - [L/V]dI/dt) ===> time constant = (V/[I x L])

  Since the battery's emf, V, is assumed to be constant, as I builds up the rate at which it builds up slows down, that is, again (as for RC circuits), the current initially increases rapidly and then slowly approaches its steady state value of I = V/R.

  This is again exponential growth where the time constant is now given by

  time constant = L / R