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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79

Problem 21 Medium Difficulty

The hot resistance of a flashlight bulb is $2.30 \Omega,$ and it is run by a $1.58-$ V alkaline cell having a $0.100-\Omega$ internal resistance. (a) What current flows? (b) Calculate the power supplied to the bulb using $I^{2} R_{\text { bulb }} \cdot($ c) Is this power the same as calculated using $\frac{V^{2}}{R_{\text { bulb }}} ?$

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Physics 102 Electricity and Magnetism

College Physics for AP® Courses

Chapter 21

Circuits, Bioelectricity, and DC Instruments

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Problem 11
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Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
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Problem 24
Problem 25
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Problem 31
Problem 32
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Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
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Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
Problem 60
Problem 61
Problem 62
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Problem 75
Problem 76
Problem 77
Problem 78
Problem 79

Video Transcript

So in the first part we're supposed to find the current. So we know that when we connect the battery like this. So this is the bike here, which has internal resistance of two point feeble, and this is internal resistance. This is resistance off 2.3. I should not call it internal, but this is external assistance of note resistance, so we can see that resistance more are on Capitol. Are are are in serious. So this is the M f E. So even if he is being distributed alone, smaller on the capital are and the current is decided accordingly, so current will be in if divided by uppercase off. Just lower case R on Let's substitute the values we know 1.58 divided by 2.3 plus 0.1. This is 0.1, so this will be 2.4. If you saw this, you'll find the answer is 0.65 athe, so we'll keep only three significant digits appears question. BCE's calculate the power supply to the bulb using I square are off a butt so power supply to the bulb is equal to the square are off about this will be simply 0.6583 squared times resistance off about it, which is 2.3. This is equal to 0.99 69 Since we want only three things that indicates this must be 0.997 what in part C questions is. What if we used the formula We squared, divided by capital up, which is going to hell, the same value which is supposed to hear the same value. But in this case, let's see, let's calculate it so I can explain it better. So the if you use one point fine eight squared, divided by the resistance off 2.3. So that is 1.5 times 158 divided by 2 to 3. You will get the power as one point 08 which is not consistent with this. But we have to understand that we cannot use enough here directly, so we must We must first find the value off B, which is equal to Yemen minus all times more art, which is 1.5 wait minus. The current was 0.658 The times 0.1. So let me do this one. So one point by eight minus 0.65 weeks times 0.1, This is equal to one point 51 for two volt. So now when we must use this value here. So let's do that. Peeing is equal to one point. If I won squared, divided by 2.3, this is equal to zero point 99 seven. What? As you can see now we should we do get we do get the same value with the ball.

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