The current in a wire varies with time according to the relationship I = 55 A - (0.65 A/s^2)t^2. (b) What constant current would transport the same charge in the same time interval?

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Everyone in this problem. An appliance is subjected to a current that follows the current time relationship. 22 A 22 amps minus 0.21 amps per second squared times T squared. And were asked to determine the constant current That can transport the same charge within 10 seconds. Yeah. Alright so we're looking for information about the current. Now let's recall that the current I. Is equal to D. Q. DT. Okay we're asked to find a current with the same charge. So we want some information about charge. Some information about current. This equation relates the two. Okay because we have the change in the charge over time is equal to the current. I. Alright so if we can figure out what charge this current that isn't constant transports then we can try to figure out what constant current we need. Okay so first things first we need to figure out what is this charge that we're transporting. So if we rearrange this equation we can write that D. Q. Is equal to I. D. T. Okay we can integrate both sides on the left we're gonna have Q. And on the right we're gonna have the integral of I. D. T. Alright so using the information we're given, We want to do this within 10 seconds. Okay so we're gonna start from zero All the way up to 10 seconds. Okay and our current I. Is given by this relationship. So we're going to integrate this relationship 22 amps minus 0.21 amps per second squared times T squared D. T. So our charge is given by the integral of our current. Our current is given by this relationship. 22 amps minus 220.21 amps per second squared t squared. Ok, so we're going to integrate that. We want to figure this out within 10 seconds. So we're integrating from 0 to 10 seconds. All right. So if we integrate okay, we have 22 amps with respect to D. T. This is just gonna be 22 Amps Times T Okay. And then we have 0.21 T squared integrated with respect to T is going to be 0.21 amps per second squared And then you're going to get t cubed over three. Okay, so we raised the exponent by one and divide by that exponent and all of this is going to be evaluated from T equals zero seconds to t equals 10 seconds. Alright, so at 10 seconds we're gonna have 22 amps times 10 seconds minus 0.21 amps per second squared times 10 seconds. Cute. All over three. And then we're going to subtract when we evaluate at zero but when we evaluate at zero we just get zero in the first term, zero in the second term. So we just subtract 20. so we don't really need to worry about that. Alright, so this first term is going to be 220 seconds And the second term works out to 70 amp seconds K Because we have an amp per second squared times second. Q. So we get amped second. This is going to give us 150 amp seconds. Okay? And this is the equivalent unit to the Cool. Um so we have 100 and 50. Cool. Alright, so we have our charge of 150 columns. Okay, that is the charge that we are transporting and we want to know instead of having this current that changes with time what constant current will give us the same charge? Well, let's just go up for a second. Okay, we have this relationship between the current and the charge. So, if we're looking for our current, I We know that the charge we want is columns. We know that this is within 10 seconds. So we have 1 50 columns divided by 10 seconds, which is gonna give us a current of 15 amperes. All right. So, if we look at our answer choices, okay? We see that we have answer choice C. Okay. The constant current that can transport the same charge is going to be amperes. Thanks everyone for watching. I hope this video helped see you in the next one

The current in a wire varies with time according to the relationship I = 55 A - (0.65 A/s^2)t^2. (b) What constant current would transport the same charge in the same time interval?

Verified Solution

Key Concepts

Current (I)

Charge (Q)

Integration