Video: Calculating Instantaneous Values of a Time-Dependent Current

The charge that flows through a point in a wire as a function of time is modelled as π‘ž(𝑑) = π‘žβ‚€π‘’^(βˆ’π‘‘/𝑇), where π‘žβ‚€ = 5.0 c and capital 𝑇 = 5.0 s. What is the initial current 𝐼₀ through the wire at a time 𝑑 = 0.00 s? Find the magnitude of the current at the instant 𝑑 = 0.25𝑇. At what value of 𝑑 will the current be reduced to exactly 2/3 𝐼₀?

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Video Transcript

The charge that flows through a point in a wire as a function of time is modelled as π‘ž as a function of 𝑑 is equal to π‘ž sub zero times 𝑒 to the negative 𝑑 over capital 𝑇, where π‘ž sub zero equals 5.0 coulombs and capital 𝑇 equals 5.0 seconds. What is the initial current 𝐼 sub zero through the wire at a time 𝑑 equals 0.00 seconds? Find the magnitude of the current at the instant 𝑑 equals 0.25 times capital 𝑇. At what value of 𝑑 will the current be reduced to exactly two-thirds 𝐼 sub zero?

Let’s begin with the first of these three questions, which is to solve for the initial current, we’re calling it 𝐼 sub zero, through the wire when 𝑑 is equal to 0.00 seconds. To solve for this value 𝐼 sub zero, we’ll want to remember a connection between charge 𝑄 and current 𝐼. Current 𝐼 is equal to the amount of charge that flows past a given point in some amount of time capital 𝑇; that is, the current in a circuit is equal to the time rate of change of charge in that circuit. One way to write this is to use Ξ” symbols to say that 𝐼 is equal to the change in charge divided by the change in time. And as this change becomes infinitesimally small, we can adopt a notation more familiar in calculus and say that 𝐼 is equal to 𝑑𝑄 over 𝑑𝑇.

And then looking at what’s given to us in the problem statement, we see we have an expression for charge as a function of time 𝑑. This means that if we wanna solve for the current as a function of time through this point in a circuit, then that’s equal to the time derivative of π‘ž as a function of 𝑑. Snd since that function is equal to π‘ž sub zero, the initial amount of charge, times 𝑒 to the negative 𝑑 over capital 𝑇, we can plug that expression in for π‘ž of 𝑑 here.

There we have it! And now we see your next step is to take the time derivative of this expression. The time derivative of π‘ž sub zero times 𝑒 to the negative 𝑑 over capital 𝑇. Well, remember capital 𝑇 is a constant value: 5.0 seconds. If we factor out the constant, π‘ž sub zero, we see our real task is to take the time derivative of 𝑒 to the negative 𝑑 over capital 𝑇. And by the way the exponential function works, that derivative is equal to the original term itself 𝑒 to the negative 𝑑 over capital 𝑇 multiplied by the derivative of the exponent, and that derivative is negative one over capital 𝑇.

Another way to write this would be as negative π‘ž sub Zero over capital 𝑇 multiplied by 𝑒 to the negative 𝑑 over capital 𝑇. What we’ve solved for here is current as a function of time; that is, we can plug in anytime value for lowercase 𝑑 and solve for the current at that particular time. That’s what we want to do here. We want to solve for the initial current, which occurs at a time value of 0.00 seconds. 𝐼 sub zero then is equal to negative π‘ž sub zero over capital 𝑇 times 𝑒 to the negative 0.00 seconds over capital 𝑇. Looking at this base, which we’ll raise into the zeroth power, we know that anytime we take a base to the zeroth power, that’s equal to one.

So our expression simplifies somewhat: 𝐼 sub zero is equal to negative π‘ž sub zero over 𝑑. And π‘ž sub zero we’re told is 5.0 coulombs, and capital 𝑇 is 5.0 seconds. Before we calculate this fraction though, let’s have a word about this minus sign. We know that when we consider current flow in a circuit, we conventionally consider current to flow in the direction of positive charge motion, understanding that it’s not the positive charges but the negative charges, the electrons, that actually do the moving. With that understanding of the current in our scenario, 𝐼 sub zero, we can take the negative of this value and call that the current.

And that’s because the only difference between these two currents is what type of charge we’re considering to be in motion. When we leave off the minus sign then, we find that 𝐼 sub zero is equal to π‘ž sub zero over capital 𝑇, which is equal to 5.0 coulombs divided by 5.0 seconds. And this is equal to 1.0 coulomb per second or 1.0 amperes. That’s the current flowing through this point in the wire at a time value of 0.00 seconds. Now let’s consider our second question, which has to do with the current and the circuit at a different time value. In this part of the question, we want to solve for the magnitude of the current in the circuit when 𝑑, the time, is equal to 0.25 times capital 𝑇, recalling the capital 𝑇 has a value of 5.0 seconds.

As we begin let’s recall our previous solution for the current in a circuit 𝐼 as a function of 𝑑. Earlier we found that current at this point as a function of time is equal to negative π‘ž sub zero over capital 𝑇 times 𝑒 to the negative 𝑑 over capital 𝑇. And recall further that lowercase 𝑑 is a variable in this function, and uppercase 𝑇 is a constant as is π‘ž sub zero. In part one, we solved for this current when time was equal to 0.00 seconds. Now we want to do something similar, but for a different time, when time is equal to 0.25 times capital 𝑇.

To solve for this current value, we can plug in 0.25 times capital 𝑇 for lowercase 𝑑 in this expression. When we make this substitution, notice we’ve written the exponential function a bit differently so it’s argument is more legible. Now that we’ve done this, we have an expression for the current in our circuit when time is equal to 0.25 times capital 𝑇. And when we look at the argument of our exponential function, we see it’s negative 0.25 times capital 𝑇 divided by 𝑇. What happens to the two capital 𝑇s? They cancel one another out. We have then this simplified expression for the current at this particular time value. To solve for it, all we’ll want to do now is to plug in for π‘ž sub zero and capital 𝑇.

And remember that those values are given to us in the problem statement. Before we solve for this current value, there’s one last step to take. Recall that we wanna find the magnitude of the current at this particular time. So when we put absolute value bars around the left-hand side of our equation, that means that we get rid of the minus sign on the right-hand side. We don’t have to change the sign in the argument of our exponential function because, as we’ll see, 𝑒 to the negative 0.25 is a positive value. When we calculate this expression, we find it’s equal to 0.78 coulombs per second or 0.78 amps. That’s how much current is flowing through this point in the wire at a time value of 0.25 times capital 𝑇.

Let’s now turn to the final part of our question. In this last part of the question, we want to solve for the time value 𝑑 at which the current in the circuit is equal to two-thirds the initial current in the circuit 𝐼 sub zero that we solved for in part one. In other words, at what time does the current in the circuit equals two-thirds of one amp? To find that out, we’ll go back to our original solution for 𝐼 current as a function of 𝑑. Here is our now somewhat familiar expression for the current in the circuit as a function of time. What we want to do in this last step is to rearrange this entire expression so we have 𝑑, time, by itself on one side of the equation. When we do that, we’ll effectively have solved for 𝑑 as a function of 𝐼, whereas now we have 𝐼 as a function of 𝑑.

So looking at this equation, what steps can we take to isolate the variable 𝑑. For one thing, we can multiply both sides of the equation by negative capital 𝑇 over π‘ž sub zero. When we do that, all we have left on the right-hand side is 𝑒 to the negative 𝑑 over capital 𝑇. Now we can recall a mathematical relationship between two complementary functions. In this exercise, we’ve been working with the exponential function. And a complementary function to that is the natural logarithm. In fact, if we take the natural logarithm of 𝑒 to the π‘₯, that’s equal simply to the exponent π‘₯. That means that if we take the natural logarithm of both sides of this equation, we’ll simplify our exponent here so that the right-hand side of our expression is negative 𝑑 over capital 𝑇.

As a final step, if we multiply both sides of this expression by negative capital 𝑇, then that term will cancel out on our right-hand side and we’ll be left simply with 𝑑. Here’s what we have then: 𝑑, the time, is equal to negative capital 𝑇 times the natural logarithm of negative capital 𝑇 over π‘ž sub zero times 𝐼 as a function of 𝑑, the current in the circuit at that time. And in our case, our current value is set to be equal to two-thirds of the initial current in the circuit, 𝐼 sub zero. So we’ll make that substitution. At this point, all that’s left for us to do before we calculate is to substitute in for our values of capital 𝑇, π‘ž sub zero, and 𝐼 sub zero, all of which we’ve either been given in a problem statement or have solved for already.

Before we calculate this expression, notice in the argument of our natural logarithm function that our units are seconds per coulomb in our first term and then amperes, which are coulombs per second, in the second term. What that tells us is that all of the units within this argument of the natural logarithm function cancel out. In the end then, the unit of our final answer will simply be seconds, consistent with our expectation that it’d be a time. And when we calculate all this, to two significant figures, we find that 𝑑 is equal to 2.0 seconds. That’s the time value at which the current in our circuit is equal to two-thirds its initial value. 𝐼 sub zero.

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