Analyzing such a parallel RL circuit, like the one shown here, follows the same process as analyzing an RC series circuit. So if you are familiar with that procedure, this should be a breeze.

If your RL parallel circuit has an inductor connected with a network of resistors rather than a single resistor, you can use the same approach to analyze the circuit. But you have to find the Norton equivalent first, reducing the resistor network to a single resistor in parallel with a single current source.

## Start with the simple RL parallel circuit

Because the resistor and inductor are connected in parallel in the example, they must have the same voltage*v(t)*. The resistor current

*i*

_{R}*(t)*is based on Ohm’s law:

The element constraint for an inductor is given as

where *i(t)* is the inductor current and *L* is the inductance.

You need a changing current to generate voltage across an inductor. If the inductor current doesn’t change, there’s no inductor voltage, which implies a short circuit.

Now substitute*v(t) = Ldi(t)/dt*into Ohm’s law because you have the same voltage across the resistor and inductor:

Kirchhoff’s current law (KCL) says the incoming currents are equal to the outgoing currents at a node. Use KCL at Node A of the sample circuit to get *i*_{N}*(t) = i*_{R}*(t) =i(t)*.

Substitute *i*_{R}*(t)* into the KCL equation to give you

The RL parallel circuit is a first-order circuit because it’s described by a first-order differential equation, where the unknown variable is the inductor current *i(t)*. A circuit containing a single equivalent inductor and an equivalent resistor is a first-order circuit.

Knowing the inductor current gives you the magnetic energy stored in an inductor.

In general, the inductor current is referred to as a state variable because the inductor current describes the behavior of the circuit.

## Calculate the zero-input response for an RL parallel circuit

Here is how the RL parallel circuit is split up into two problems: the zero-input response and the zero-state response. Here, you’ll start by analyzing the zero-input response.To simplify matters, you set the input source (or forcing function) equal to 0: *i*_{N}*(t)* = 0 amps. This means no input current for all time — a big, fat zero. The first-order differential equation reduces to

For an input source of no current, the inductor current *i*_{Z}*I* is called a *zero-input response*. No external forces are acting on the circuit except for its initial state (or inductor current, in this case). The output is due to some initial inductor current *I** _{0}* at time

*t*= 0.

You make a reasonable guess at the solution (the natural exponential function!) and substitute your guess into the RL first-order differential equation. Assume the inductor current and solution to be

This is a reasonable guess because the time derivative of an exponential is also an exponential. Like a good friend, the exponential function won’t let you down when solving these differential equations.i_{ZI}(t) = Be^{kt}

You determine the constants *B* and *k* next. Substitute your guess *i*_{ZI}*(t)* = *Be** ^{kt}* into the differential equation:

Replacing *i*_{ZI}*(t)* with *Be** ^{kt}* and doing some math gives you the following:

You have the characteristic equation after factoring out *Be** ^{kt}*:

The characteristic equation gives you an algebraic problem to solve for the constant *k*:

Use *k = –R/L* and the initial inductor current *I** _{0}* at

*t*= 0. This implies that

*B = I*

*, so the zero-input response*

_{0}*i*

_{ZI}*(t)*gives you the following:

The constant *L/R* is called the *time constant*. The time constant provides a measure of how long an inductor current takes to go to 0 or change from one state to another.

To analyze the RL parallel circuit further, you must calculate the circuit’s zero-state response, and then add that result to the zero-input response to find the total response for the circuit.