### Video Transcript

A conductor carries a time-dependent current that decreases exponentially. The instantaneous current is modelled as πΌ equals πΌ sub zero times π to the negative π‘ over π, where π‘ corresponds to the time the current has persisted for, πΌ sub zero equals 1.50 amps is the current at π‘ equals 0.00 seconds, and the time constant π equals 0.40 seconds. How much charge flows through the conductor between π‘ equals 0.00 seconds and π‘ equals two times π?

The first thing we can take note of here is that weβre given current as a function of time, but we want to solve for a charge amount. So to get started, letβs recall a relationship between current time and charge. In general, current πΌ is equal to the amount of charge Ξπ that passes at certain point in a circuit divided by the change in time Ξπ‘.

If we shrink down these changes in charge and changes in time until theyβre infinitesimally small, using notation, we may have seen before in calculus class, we see that current is equal to ππ over ππ‘, an infinitesimal change in charge divided by an infinitesimal change in time. If we take this expression and multiply both sides of it by ππ‘, then we see that current times this infinitesimal change in time is equal to the infinitesimal change in charge ππ.

In our question though, we donβt want ππ; we want π, the total amount of charge that flows through the conductor between two time values. To solve for π rather than ππ, weβll integrate this side of the expression. And because we do it on one side, we need to integrate the other side as well.

The integral of ππ is equal to simply π. And itβs that value we want to solve for. And itβs given by the integral of the current πΌ over time between two given time values, 0.00 seconds and π‘ equals two times π.

Letβs move ahead and solve in for π by first plugging in for our current πΌ, which is given to us in our problem statement as a function of π‘. There! Weβve now plugged in for the functional form of the current πΌ. Itβs equal to the initial current value πΌ sub zero times π to the negative π‘ divided by π, which is a constant.

Currently, we see that our integral is an indefinite integral. But we wanna get a definite result for our charge π. To do that, weβll put π points on this integration. What are those π points? Theyβre the time values given to us: 0.00 seconds on the low end and two times π on the high end.

With all this done, weβre now ready to compute this integral. And as a first step, since πΌ sub zero is a constant value, letβs take it outside of the integral. We now see that weβre taking the integral of π to the negative π‘ over π. And we recall that the integral of this expression will be that which if we take its derivative will give us back π to the negative π‘ divided by π.

Doing a little bit of work off to the side, we see that if we take the time derivative of the expression negative π times π to the negative π‘ divided by π, then that derivative is equal to π to the negative π‘ over π, which is the same as the argument of our integral. This tells us that the integral of π to the negative π‘ over π is indeed negative π times π to the negative π‘ divided by π.

Our next step in solving for π then is to evaluate this integral at the two endpoints, 0.00 seconds and two times π. We plug those two values in for π‘ in our expression. Hereβs what that looks like. In our first term, when we plug in two π for π‘, we have π to the negative two π‘ over π, where we see the factors of π cancel out. And in our second term, where we have π to the negative 0.00 divided by π, we see we have a base to an exponent of zero. And anytime we have a base to an exponent of zero, that whole term is equal to one.

The whole expression inside the parentheses then simplifies to π to the negative two minus one or written another way, we can say that the left-hand side of this expression is equal to πΌ zero times π all multiplied by the quantity one minus one over π squared. And all of that is equal to the charge π that flows through the conductor between these two given times.

Looking at this expression, weβre given a value for πΌ sub zero, thatβs 1.50, amps and for π, thatβs 0.40 seconds. So weβre ready to plug in and solve for π. As we do, remember that π represents the exponential function and that π squared is itself a pure number.

When we plug in for these values, notice that our units of seconds cancel out and weβll be left with units of coulombs that is the units of charge. And when we multiply these three terms together, we find a result of 0.52 coulombs. Thatβs how much charge flows through the conductor between π‘ equals 0.00 seconds and π‘ equals two π.