# Video: Calculating the Charge Carried by a Time-Dependent Current

A conductor carries a time-dependent current that decreases exponentially. The instantaneous current is modelled as πΌ = πΌβπ^(βπ‘/π), where π‘ corresponds to the time the current has persisted for, πΌβ = 1.50 A is the current at π‘ = 0.00 s, and the time constant π = 0.40 s. How much charge flows through the conductor between π‘ = 0.00 s and π‘ = 2π?

04:52

### Video Transcript

A conductor carries a time-dependent current that decreases exponentially. The instantaneous current is modelled as πΌ equals πΌ sub zero times π to the negative π‘ over π, where π‘ corresponds to the time the current has persisted for, πΌ 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 weβre given current as a function of time, but we want to solve for a charge amount. So to get started, letβs recall a relationship between current time and charge. In general, current πΌ is equal to the amount of charge Ξπ that passes at certain point in a circuit divided by the change in time Ξπ‘.

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

In our question though, we donβt want ππ; we want π, the total amount of charge that flows through the conductor between two time values. To solve for π rather than ππ, weβll 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 ππ is equal to simply π. And itβs that value we want to solve for. And itβs given by the integral of the current πΌ over time between two given time values, 0.00 seconds and π‘ equals two times π.

Letβs move ahead and solve in for π by first plugging in for our current πΌ, which is given to us in our problem statement as a function of π‘. There! Weβve now plugged in for the functional form of the current πΌ. Itβs equal to the initial current value πΌ 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, weβll put π points on this integration. What are those π points? Theyβre 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, weβre now ready to compute this integral. And as a first step, since πΌ sub zero is a constant value, letβs take it outside of the integral. We now see that weβre 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. Hereβs 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 πΌ 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, weβre given a value for πΌ sub zero, thatβs 1.50, amps and for π, thatβs 0.40 seconds. So weβre 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 weβll 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. Thatβs how much charge flows through the conductor between π‘ equals 0.00 seconds and π‘ equals two π.