# RL Circuits

Following on from our discussions of inductors we come to RL Circuits aka Resistor-Inductor circuits which are circuits that consist of resistors, inductors and a power source.

This will be our second practical example of a transient circuit where we will see the output current ramp up in time until reaching a steady-state value or decrease in time asymptoting to zero depending on the circuit configuration.

## The RL Charging Circuit:

The circuit shown below is the classic example of a series RL circuit where the current through the inductor ramps up once the switch is closed until reaching a steady state current:

As the switch is closed current begins to flow from the voltage source and the inductor initially induces a large back emf to resist the current with most of the voltage appearing across the inductor which decreases with time while the current increases with a typical curve as shown below which is exponential in nature approaching a steady state current equal to the voltage of the voltage source divided by the circuit resistance (Ohm's Law again :) ):

The reason why the current asymptotes to the voltage source divided by the resistance is that the voltage across the inductor approaches 0V meaning the entire source voltage is dropped across the resistor and then Ohm's Law can be applied to determine the current.

It can be shown with calculus that the current through the inductor with time for the circuit above is given by:

Where iL denotes the current flowing through the inductor, V is the voltage, t is time, R is the circuit resistance and L is the inductance.

## Time Constant:

Like RC circuits, RL circuits have a time constant. It denotes the time in seconds taken for the inductor current to rise to 63% of the source current.

The time constant for RL circuits is given by:

Thinking about this intuitively, if the inductance is higher the inductor will resist changes in current more strongly taking longer for the current to reach the steady state value. From the resistance perspective recall that the current asymptotes to the voltage source divided by the resistance and so a lower value will lead to a larger asymptote which will take longer to reach.

Then the above equation can be reduced to:

## Examples:

1. Determine the time constant for the circuit below:

2. Determine the time constant for the circuit below:

To determine the time constant for the circuit above we need to calculate the total resistance and inductance in the circuit. The total resistance is simply R1||R2 but what is the total inductance? Recall that inductance adds like resistance, we have L1 in parallel with L2 and then L3 in series with them so the total inductance is equal to L1||L2 + L3. Then the time constant for this circuit is given by:

3. a) Determine the current through the inductor in the circuit below for time t after the circuit has been switched on:

Recall that the equation for the current through an inductor is given by: The first step is to determine the time constant for the circuit by considering the circuit once the switch has been flicked on. In this case it is simply L divided by R: We can then substitute these into the formula to calculate the current through the inductor for time t:

b) Sketch the voltage vs time curve:

To sketch the curve we can calculate the times for 1, 2, 3, 4 and 5 time constants and then equate the corresponding voltages and connect the points with an exponential curve approximation: We can present these in a table:

## The RL Discharging Circuit:

The circuit shown below is the classic example of an RL discharging circuit where the inductor acts as a current source with current decreasing exponentially with time as the switch is closed with the magnetic field eventually collapsing completely:

If we assume the inductor has been charged (stored energy in a magnetic field), then once the switch is closed current will continue to flow in the same direction (to try to retain the previous current) and the discharge curve will look as shown below:

It can be shown with calculus that the current supplied by the inductor with time for the circuit above is given by:

Where IL denotes the charged current through the inductor, t is time, and τ is the time constant.

For the discharging circuit shown, the time constant is 2us and the charged current is 1mA hence the current through the inductor is given by:

## Questions:

1. a) Determine the time constant for the circuit below:

20ns

b) Determine the equation for the current through the inductor once the switch has been closed.

0.001(1-e^(-50*10^6t))

c) Sketch the current vs time curve for the current flowing through the inductor.

d) Determine the current through the inductor once the switch has been closed for time t.

993uA

e) Determine the current flowing through the inductor after 1 time constant.

632uA

2. a) Determine the time constant for the circuit below:

9.375ns

b) Determine the equation for the current through the inductor once the switch has been closed.

750*10^-6(1-e^-(106666666.7t))

c) Determine the current flowing through the inductor once the switch has been closed for time t.

500uA

3. a) Determine the time constant for the circuit below assuming the equivalent inductance had a current of 20mA flowing through it.

6.184ns

b) Determine the equation for the current through the inductor once the switch has been closed.

20*10^-3(e^-(161707632.6t))

c) Sketch the current vs time curve for the inductor.

e) Determine the current flowing through the inductor after 1 time constant.