The current in a wire varies with time according to the relationship I = 55 A - (0.65 A/s^2)t^2 . (a) How many coulombs of charge pass a cross section of the wire in the time interval between t = 0 and t = 8.0 s? (b) What constant current would transport the same charge in the same time interval?

Video transcript

Hey everyone in this problem. The electric current flowing through an element in a circuit follows the relation amps minus 300.25 amps per second squared times T squared. And we're asked to determine the Coolum of charge That passes a fixed point in the element within 12 seconds. Alright. So we're given the electric current and recall that electric current can be written as I we know that that's equal to 30 amps minus 0. amps per second squared times T squared. Okay. And so our current depends on the time. T. And recall it. The current I is equal to the change in the charge over the change in time, multiplying both sides by D. T. We get that the change in the charge D. Q. Is equal to the current. I there's a change in time DT and we can integrate both sides in order to get the charge Q. When we get the integral of I. D. T. On the right hand side. So we're looking for the cool um of charge. We're looking for this value of Q. And now we have a way to calculate it. So our integral OK We're looking for within seconds. Okay, so we're gonna integrate from zero seconds All the way up to seconds. That's going to give us the amount of charge that passes through that point in 12 seconds. Within those 12 seconds of our current I which is 30 amps minus 0.25 amps per second squared T squared D. T. All right, let's move down. And now we need to integrate this. We have a polynomial here in terms of T. So it's not too bad to integrate. Okay, we're gonna get the integral of 30 a amps is just gonna be 30 times T. So we get 30 amps times the time T and then we have the integral of some constant times, T squared. The integral of T squared is going to be t cute over three. Can you raise the exponent by one, divide by the new exponent. So we get minus 0.25 amps per second square times T cubed over three. And all of this is going to be evaluated from zero seconds to seconds. Alright, so starting with 12 seconds we have amps times 12 seconds -0.25 amps per second squared times 12 seconds cute over three. And then we're going to subtract the evaluation at zero seconds. Now when we evaluate at zero seconds 30 amps times zero gives us zero and 0.25 amps per second square time. Zero cubed over three gives us zero as well. So we just get minus zero there Now 30 amps times 12 seconds. That's going to give us 360 and seconds 0.25 amp per second squared times seconds cubed over three. That's going to give us 144. Again, the unit is amp second because we have amp and then we have divided by seconds squared times second cubed. And this is gonna give us a value of 216 amps times seconds, Which is the unit of cool. And so we get 216 cool. Oh, that is going to be the charge that passes through a fixed point in the element within 12 seconds. And so if we go back up to our answer traces, we see that the answer is going to be a that's it for this one. Thanks everyone for watching. See you in the next video.