# Differential EQ of this RL Circuit

In summary: Hello saad87,Did you solve the differential equation? If so, what was the answer? In summation, did you find the voltage across the second resistor?
Hi,

I have the above the circuit and I've ended up with the following differential equation. I've tried solving it by separating the variables, but am not sure if its the right answer.

IR + L(di/dt) + IR = x(t)

x(t) = 5V. But since IR = y(t) and di/dt = 1/R dy/dt.

y(t) is the Voltage across the 2nd Resistor.

y(t) + L/2R(dy/dt) = 2.5

Solution:

y(t) = 2.5 - e(-2*R*t)/L

y(t) is the voltage across the 2nd Resistor.

I've tried simulating it using an online applet, and I've not been getting the predicted result. Any help? How do I analyze this?

Nevermind, just solved it.

The general solution to that equation is $y= 2.5+ Ce^{-2Rt/L}$ but to determine C you need more information, say the value of y when t= 0. With C= 1, y(0)= 2.5- 1= 1.5 volts. Is that true?

Hi,

I have the above the circuit and I've ended up with the following differential equation. I've tried solving it by separating the variables, but am not sure if its the right answer.

IR + L(di/dt) + IR = x(t)

x(t) = 5V. But since IR = y(t) and di/dt = 1/R dy/dt.

y(t) is the Voltage across the 2nd Resistor.

y(t) + L/2R(dy/dt) = 2.5

Solution:

y(t) = 2.5 - e(-2*R*t)/L

y(t) is the voltage across the 2nd Resistor.

I've tried simulating it using an online applet, and I've not been getting the predicted result. Any help? How do I analyze this?

Hello saad87, welcome to the forum.

Your original equation is OK. The way you solve it is not OK though. If you rewrite the original equation as:
$$2\cdot R \cdot i + L\cdot \frac{di}{dt}=x(t)$$
or
$$\frac{di}{dt}=5-200\cdot i$$
Can you proceed in solving this for i?

After having found i you can find the voltage across the resisters by applying the law of Ohm and the voltage across the coil with:
$$L\cdot \frac{di}{dt}$$

coomast

Oeps, too late...

Yeah, the only thing I was doing wrong I was not applying the limits, at all. Smacked my head when I found what was wrong.

Anyway, here's what I"m confused about. I'm studying a course on Signals & Systems, and the above question confused me because there were no initial conditions specified!

It simply said "find the response of the system if x(t) = 0 for t<0 and x(t)=5 for t>0"

Does this mean, that since x = 0 for t<0, that I can assume y(0) = 0?

What you need to do is to use the law of Ohm to find the voltage across the second resistor. This is y(t)=R*i(t)=100*i(t).
The thing to do is thus to find the current or the solution to the equation I described. The initial condition is i(0)=0. This because the voltage is applied at t=0, and at this time no current is flowing because of the presence of the coil. The complete voltage at the beginning is across this coil and goes down for t>0. This is the standard behaviour of an series-LR circuit.
Did you find the current meaning did you solve the differential equation?

coomast

coomast said:

What you need to do is to use the law of Ohm to find the voltage across the second resistor. This is y(t)=R*i(t)=100*i(t).
The thing to do is thus to find the current or the solution to the equation I described. The initial condition is i(0)=0. This because the voltage is applied at t=0, and at this time no current is flowing because of the presence of the coil. The complete voltage at the beginning is across this coil and goes down for t>0. This is the standard behaviour of an series-LR circuit.
Did you find the current meaning did you solve the differential equation?

coomast

Yeah, I only wanted to know if v(0) = 0. I didn't find the current and instead, found out the voltage on the 2nd Resistor directly using the differential equation I mentioned above.

I then simulated the circuit and found the voltage across the resistor at a certain time T (5 ms if you must know), and the results from my equation and the simulation agree. So I'm satisfied that my answer is correct!

## 1. What is a differential equation?

A differential equation is a mathematical equation that relates a function to its derivatives. It represents the rate of change of a system and is commonly used in physics, engineering, and other scientific fields.

## 2. What is an RL circuit?

An RL circuit is a type of electrical circuit that contains a resistor (R) and an inductor (L). The inductor stores energy in the form of a magnetic field, while the resistor dissipates energy in the form of heat. This type of circuit is commonly used in electronic devices.

## 3. What is the differential equation for an RL circuit?

The differential equation for an RL circuit is given by dI/dt + RI = V, where I is the current flowing through the circuit, R is the resistance, V is the applied voltage, and t is time. This equation represents the relationship between the rate of change of current and the applied voltage in the circuit.

## 4. How do you solve a differential equation for an RL circuit?

To solve a differential equation for an RL circuit, you can use techniques such as separation of variables, substitution, or integrating factors. It is also possible to use software programs or online solvers to obtain a numerical solution.

## 5. What are the applications of differential equations in RL circuits?

Differential equations are used to model and analyze the behavior of RL circuits in various applications, such as power systems, electric motors, and electronic filters. They allow engineers and scientists to predict the behavior of these circuits and make design decisions to optimize their performance.

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