Finding the current for each time interval (Electric circuits)

[s3a]

Homework Statement

The problem and its solution are attached as TheProblemAndSolution.jpg.

2. Relevant equation
ν = L di/dt

The Attempt at a Solution

I don't understand how the solution gets i = 10.0 A for t = 2 ms as well as to how it gets i = 30 – (10 * 10^3) t) (A) for 2 ms < t < 4 ms. More specifically, I see that the equation when 0 < t < 2 ms has i = 10.0 A if one plugs in 2 ms in the t variable like so, 5 * 10^3 (2 * 10^(-3) s) = 10.0 A. However, what I do not get is, why does the equation for the interval (0, 2) work for the point t = 2 ms (does the reasoning relate to the fact that 1.9999 (periodic 9) can be proved to equal 2? As to when 2 < t < 4, something seems off with the first equality but I cannot figure out exactly what. Just to note, I don't see anything wrong with Fig. 2-12 and the way it relates to the equations in the work so I don't think anything is wrong but, to reiterate, I don't get why i = 10.0 A at t = 2 ms and I'm confused as to how to get the equation for when 2 ms < t < 4 ms because the first equality is messed up which throws me off for the rest of the work.

Any help in understanding these things would be greatly appreciated!

 

[TSny]

You have the right idea about the current at t = 2 ms. The current will be a continuous function of time, so the value at t = 2 ms is the limit of the expression for the first time interval as t approaches 2 ms.

There is a typo in the first line of the expression for i in the second time interval. It should read something like i = 10 + ∫vdt = 10 + (1/.003) ∫-30dt . The second and third lines look ok.

 

[s3a]

I'm still having doubts as to whether (t = 2 ms, i = 10.0 A) should be an included point or not.

Is it correct to say that it is BECAUSE equating the 0 < t < 2 ms and 2 ms < t < 4 ms equations as follows will yield an input t such that each equation's output is i = 10.0 A?:

5 * 10^3 t = 30 - 10 * 10^3 t
15 * 10^3 t = 30
t = 30/(15*10^3) [Plugging this value in either equation yields i = 10.0 A.]

 

[TSny]

I think of it this way. For a circuit that has inductance, the current cannot change discontinuously even when the voltage changes discontinuously. [To change the current discontinuously would require a "dirac-delta" type of impulsive voltage which is rather fictitious and is not what you are dealing with here.] So, the current will be a continuous function of time. So, the current at t = 2 ms must be the limit as t approaches 2 ms in the interval 0 < t < 2 ms.

From calculus, we know that if you take the integral from 0 to t of a function with a finite discontinuity, then the integral will be a continuous function of t. Here, the current at any time t is the integral of V from 0 to t. Even though V is discontinuous at t = 2 ms, the current will be continuous at t = 2 ms.

 

[Nickie]

The solution provided is correct, but I can understand why it may be confusing. The key to understanding this problem is to remember that the current is not constant in an RL circuit, but rather changes over time. In order to find the current at any given time, we need to use the equation ν = L di/dt and solve for i.

For the interval 0 < t < 2 ms, we are given the inductance (L) as 5 * 10^3 H and the rate of change of current (di/dt) as 10 A/ms. Plugging these values into the equation, we get ν = (5 * 10^3 H)(10 A/ms) = 50 V. This voltage is constant throughout the interval, so we can use Ohm's Law to find the current, which is i = V/R = 50V/5Ω = 10 A.

For the interval 2 ms < t < 4 ms, we need to take into account that the current is changing over time. In this case, we can use the initial current (i = 10 A) at t = 2 ms and the rate of change of current (di/dt = -10 A/ms) to find the equation for current. This is done by integrating the rate of change of current over the time interval, which gives us i = 10 A - (10 A/ms)(t - 2 ms) = 30 A - (10 A/ms)t. This equation gives us the current at any given time within the interval.

I hope this helps clarify the solution for you. Remember to always pay attention to the units and how they are used in the equations, as this can help in understanding the solution better.