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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

Chapter 21

Circuits, Bioelectricity, and DC Instruments

Direct-Current Circuits

University of Michigan - Ann Arbor

University of Sheffield

University of Winnipeg

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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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