5. Application of Ordinary Differential Equations: Series RL Circuit

Series RL circuit diagram
RL circuit diagram

The RL circuit shown above has a resistor and an inductor connected in series. A constant voltage V is applied when the switch is closed.

The (variable) voltage across the resistor is given by:

`V_R=iR`

The (variable) voltage across the inductor is given by:

`V_L=L(di)/(dt)`

Kirchhoff's voltage law says that the directed sum of the voltages around a circuit must be zero. This results in the following differential equation:

`Ri+L(di)/(dt)=V`

Once the switch is closed, the current in the circuit is not constant. Instead, it will build up from zero to some steady state.

Continues below

Solving the DE for a Series RL Circuit

The solution of the differential equation `Ri+L(di)/(dt)=V` is:

`i=V/R(1-e^(-(R"/"L)t))`

[We did the same problem but with particular values back in section 2. Separation of Variables]

Here is the graph of this equation:

Graph of i=V/R(1-e^(-(R/L)t)), for a series RL circuit - solution of ordinary DE t i τ
`V/R`

Graph of `i=V/R(1-e^(-(R"/"L)t))`.

The plot shows the transition period during which the current adjusts from its initial value of zero to the final value `V/R`, which is the steady state.

The Time Constant

The time constant (TC), known as τ, of the function

`i=V/R(1-e^(-(R"/"L)t))`

is the time at which `R/L` is unity ( = 1). Thus for the RL transient, the time constant is `\tau = L/R` seconds.

NOTE: τ is the Greek letter "tau" and is not the same as T or the time variable t, even though it looks very similar.

At 1 τ

`1-e^(-(R"/"L)t)`

`=1-e^-1`

`=1-0.368`

`=0.632`

At this time the current is 63.2% of its final value.

Similarly at 2 τ,

`1 - e^-2= 1 - 0.135 = 0.865`

The current is 86.5% of its final value.

After 5 τ the transient is generally regarded as terminated. For convenience, the time constant τ is the unit used to plot the current of the equation

`i=V/R(1-e^(-(R"/"L)t))`

That is, since `tau=L/R`, we think of it as:

`i=V/R(1-e^(-t"/"\tau))`

Let's now look at some examples of RL circuits.

Example 1

An RL circuit has an emf of 5 V, a resistance of 50 Ω, an inductance of 1 H, and no initial current.

Find the current in the circuit at any time t. Distinguish between the transient and steady-state current.

Example 2

A series RL circuit with R = 50 Ω and L = 10 H has a constant voltage V = 100 V applied at t = 0 by the closing of a switch.

Find

(a) the equation for i (you may use the formula rather than DE),

(b) the current at t = 0.5 s

(c) the expressions for VR and VL

(d) the time at which VR = VL

Two-mesh Circuits

The next two examples are "two-mesh" types where the differential equations become more sophisticated. We will use Scientific Notebook to do the grunt work once we have set up the correct equations.

Example 3

In the two-mesh network shown below, the switch is closed at t = 0 and the voltage source is given by V = 150 sin 1000t V. Find the mesh currents i1 and i2 as given in the diagram.

2 mesh circuit diagram

Example 4

The switch is closed at t = 0 in the two-mesh network shown below. The voltage source is given by V = 30 sin 100t V. Find the mesh currents i1 and i2 as given in the diagram.

2 mesh circuit diagram