# RL circuit – detailed mathematical analysis

In this tutorial we are going to perform a very detailed mathematical analysis of a RL circuit. By the end of the article the reader will be able to understand how the current response of an RL circuit is calculated and how the principle of superposition is applied in practice.

An RL circuit is quite common in any electric machine. The winding of an electric machine (motor or generator) is represented as a RL circuit. Having a deep knowledge of the mathematical apparatus of a RL circuit will help with the understanding of the phenomena happening in the winding (coils) of an electric machine.

Let’s consider the following circuit where a continuous voltage source is connected in series with a switch, resistor and inductor. The switch has the role of opening and closing the electric circuit. When the switch is open, the voltage source is disconnected from the rest of the circuit.

Image: Series RL circuit schematic

where:

E [V] – continuous voltage source
S – switch
R [Ω] – resistance
L [H] – inductance
u [V] – voltage drop across the circuit
uR [V] – voltage drop across the resistor
uL [V] – voltage drop across the inductor
i [A] – electrical current through the circuit

We are going to do a couple of assumptions about our series RL circuit:

• both electrical components (resistor and inductor) are linear
• there is no voltage drop across the switch S
• the resistance R and inductance L have constant values
• the internal resistance of the voltage source is zero

From the Systems Modeling tutorials we can identify that our circuit has three elements:

The voltage drop on the resistor and inductor can be written as:

$\begin{split} u_R & = Ri\\ u_L & = L \frac{di}{dt} \end{split}$

If we consider out RL circuit as a system with input and output, we’ll consider that the input is the voltage drop u(t) across the circuit and the output is the current i(t) through the circuit. Both quantities are variable in time.

When the switch S is open, the input voltage u is zero. When the switch is closed, the input voltage increases suddenly to the value E of the voltage source. We can say that the input voltage u is a step input.

Image: RL series circuit – free and forced current response

When the switch closes, the output (response) of the circuit, the electrical current i, will start to rise as described in the image above. The output current will be composed by the addition of two responses:

• a forced response iforced
• a free response ifree
$i(t) = i_{\text{forced}}(t) +i_{\text{free}}(t) \tag{1}$

The superposition principle states that for all linear systems, the final response at a given place and time caused by two or more stimuli, is the sum of the responses that would have been caused by each stimulus individually.

In our case, one of the stimulus is the input voltage u(t), which will trigger the forced response. The other stimulus is the initial condition of system, which will determine the free response.

If we keep the switch S closed for a long time the current in the circuit will stabilize at a constant value. This means that the electric circuit is in a steady (stationary) state.

If a continuous voltage is applied to an inductor, its effect will be zero. This is because the voltage drop across the inductor depends on the variation of the current. With a constant voltage, we’ll have a constant current, which means no effect of the inductance.

$\frac{di}{dt}=0 \tag{2}$ $u_L = L \frac{di}{dt}= 0 \tag{3}$

From Kirchhoff’s Voltage Law (KVL) we know that, for any loop in an electrical circuit, the sum of the electrical voltage across the loop is zero. This translates in:

$u(t) – u_R (t) – u_L (t) = 0 \tag{4}$

Because we are in a steady state, we’ll have:

$\begin{split} u(t) &= E\\ u_L (t) &= 0\\ u_R (t) &= Ri \end{split}$

Replacing these in equation (4) gives the value of the forced response:

$\bbox[#FFFF9D]{i_{\text{forced}}(t) = \frac{E}{R}} \tag{5}$

To find the initial condition of the free response, we need to analyze what’s happening at the beginning of the steady state. Time t = 0 is the time exactly before the switch S is closed. Time t = 0+ is the time exactly after the switch S is closed. In this scenario, the voltage u and the current i will be:

$\begin{split} u(0_{-}) &= 0\\ u(0_{+}) &= E\\ i(0_{-}) &= 0\\ i(0_{+}) &= 0\\ \end{split}$

The combined output current i, function of the forced and free responses, at time t = 0+, will be:

$i(0_{+}) = i_{\text{forced}} (0_{+}) + i_{\text{free}} (0_{+}) = 0 \tag{6}$

Equation (6) gives the initial condition of the free response of the system:

$i_{\text{free}} (0_{+}) = – i_{\text{forced}} (0_{+}) = – \frac{E}{R} \tag{7}$

To find the equation that describes the free current response of the system, we are going to write Kirchhoff’s Voltage Law (KVL):

$\begin{split} u(t) &= u_R (t) + u_L (t)\\ E &= R i(t) + L \frac{di(t)}{dt}\\ \end{split}$

If we replace i(t) with equation (1), we get:

$L \frac{d(i_{\text{forced}} + i_{\text{free}})}{dt} + R (i_{\text{forced}} + i_{\text{free}}) = E \tag{8}$
$L \frac{d i_{\text{forced}}}{dt} + R i_{\text{forced}} + L \frac{d i_{\text{free}}}{dt} + R i_{\text{free}}= E \tag{9}$

The forced current response is constant, which makes:

$L \frac{d i_{\text{forced}}}{dt} = 0 \tag{10}$

Also, from equation (5), we know that:

$R i_{\text{forced}} = E \tag{11}$

Replacing (10) and (11) in (9) gives:

$L \frac{d i_{\text{free}}}{dt} +R i_{\text{free}} = 0 \tag{12}$

From equation (12) we can see that the free response does not depend on the voltage source E. A solution of the differential equation could be:

$i_{\text{free}} (t) = I_{\text{free}} e^{pt} \tag{13}$

The initial value of the free response is:

$i_{\text{free}} (0) = I_{\text{free}} e^{0} =I_{\text{free}} = – \frac{E}{R} \tag{14}$

The differential of the solution (13) is:

$\frac{d i_{\text{free}}}{dt} = I_{\text{free}} p e^{pt} \tag{15}$

Replacing the solution (13) and its derivative (15) in equation (12) gives:

$L I_{\text{free}} p e^{pt} + R I_{\text{free}} e^{pt} = 0 \tag{16}$

or

$(Lp + R) I_{\text{free}} e^{pt} = 0 \tag{17}$

Since the time t is zero or positive, the term ept can not be zero. This means that the only solution for the equation (17) can be found from:

$(Lp + R) = 0 \tag{18}$

The equation (18) is called the characteristic equation of the R, L elements. The solution of the equation is:

$p = – \frac{R}{L} \tag{19}$

Replacing (14) and (19) in (13) we get the general solution of the free response:

$i_{\text{free}} (t) =- \frac{E}{R} e^{- \frac{R}{L} t} \tag{20}$

The time constant T [s] of our RL circuit is given by:

$T = \frac{L}{R} \tag{21}$

The final expression of the free response is:

$\bbox[#FFFF9D]{i_{\text{free}} (t) =- \frac{E}{R} e^{- \frac{t}{T}}} \tag{22}$

Replacing equation (5) and (22) in (1) gives the combined current response of a series RL circuit for a step input:

$\bbox[#FFFF9D]{i(t) = \frac{E}{R}- \frac{E}{R} e^{- \frac{t}{T}}} \tag{23}$

In order to visualize the response of a series RL circuit, for a step voltage input, we are going to use the following Scilab script:

// Circuit parameters
E = 12;
R = 0.3;
L = 0.04;
T = L/R;
t = 0:0.01:1;

// Forced response
i_forced = E/R;

// Free and combined response
for k=1:length(t)
i_free(k) = - (E/R)*exp(-t(k)/T);
i(k) = i_forced + i_free(k);
end

// Plot of the current response at step input voltage
plot(t,ones(1,length(t))*i_forced), xgrid()
ha=gca();
ha.data_bounds=[min(t),min(i_free);max(t),i_forced+10];
plot(t,i_free,'r')
plot(t,i,'m')
legend('i_forced','i_free','i = i_forced + i_free',4)
xlabel('t [s]','FontSize',2)
ylabel('i [A]','FontSize',2)

By running the above instruction is a Scilab script file (*.sce), we get the following graphical window:

Image: RL series circuit – forced, free and combined current response

As expected, the combined current response i is made up from the superposition of the forced iforced and free ifree current responses.

In order to make sure that our calculations are correct, we are going to integrate the differential equation derived from KVL, using the ode() Scilab function.

deff('iprim=f(t,i)','iprim=(1/L)*(E-R*i)');
t0=0; tinc=0.01; tf=1; t=t0:tinc:tf;
i0=0;
i=ode(i0,t0,t,f);
plot(t,i,'r'), xgrid
xlabel('t [s]','FontSize',2)
ylabel('i [A]','FontSize',2)
title('$\Large{\frac{di(t)}{dt}=\frac{1}{L} (E-Ri(t))}$')

Running the above instruction is a Scilab script file (*.sce), we get the following graphical window:

Image: RL series circuit – ODE solve plot

As we can see we have the same results from both methods, which gives confidence in our mathematical analysis and analytical solution of the differential equation.

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