Problem

A $0.020-\Omega$ ammeter is placed in series with…

02:50

Problem 53 Hard Difficulty

A 1.00 -M\Omega voltmeter is placed in parallel with a 75.0 -k $\Omega$ resistor in a circuit. (a) Draw a circuit diagram of the connection. (b) What is the resistance of the combination? (c) If the voltage across the combination is kept the same as it was across the $75.0-\mathrm{k} \Omega$ resistor alone, what is the percent increase in current? (d) If the current through the combination is kept the same as it was through the $75.0-\mathrm{k} \Omega$ resistor alone, what is the percentage decrease in voltage? (e) Are the changes found in parts (c) and (d) significant? Discuss.

Answer

b) $R_{c o m b}=69797 \Omega$
c) $p_{I}=1.3 \cdot 10^{-11}$
d) $p_{U}=0.069$

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Physics 102 Electricity and Magnetism

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Chapter 21

Circuits, Bioelectricity, and DC Instruments

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Video Transcript

so circuit connection. For this, a circuit will be the distance of 70 popular over as Yeah, in. Apparently we have the board meter. So both the resistor and a book meter are at the same potential difference on a potential potential measurement device that his vote meter has resistance or won me over. For the part B. We can use the metre distance for president circuit for dinner, which is one divided by R. P. As you can see here, is equal to wonder about it by our one plus one. Divided by our two, I have substituted all the values already. So if you perform this calculation, you'll find the final answer is R P is equal to 7 69,067 and the answer given on the numerator page is slightly warm. Because it's a 7 69,097 it should be 7 67 for the part. See, let's use the fact that we can use old love were equal to our I. In the beginning, we only had one register, So I want is it well to be divided by one as you can see and for the final configuration, I too will be be divided for happy. Why did I do this? Because I know that potential is kept constant now. Percent increase is given by I two minus. I want divided by one times 100. Let's substitute all the values. As you can see, we can be cancer from the numerator and denominator. And before substituting the values for our one here and our PR well here in the numerator, let's simplify it better. So if you simplify this, it will look like this are going about it. The Rp minus our Wendy about it But I want and 75,000 about it by 7 69,067 minus one. So this will be equal to So after multiplying this whole bracket by 100 you will get seven point five percent. So answer given under Newman it Paige is wrong then, for the party. Question stays. If the current through the combination is kept constant so again, we will make use off one bones law. We know now current, it's kept constant, so resistance is being changed and the potential will change accordingly. So let's write that down we want is equal toe. I are one we too is equal to I R p. That's a final registration distance of combination. So or send change percent decrease will be equal to be two minus 31 I should say we want minus 32 because we choose a smaller quantity divided by V one is equal to and because we're using decrease already, so we don't want to answer the negative. So that substitute I times are one minus R p No, I did, but I I want unfortunately, I'm okay, so we can now subtract 75,000 minus 7 69,067 divided by 75,000. So let me do this one. Oh, so this is equal to six point nine percent. Thank you. And in the part e questions is are the changes find insanity significant? Well, depends how much accuracy you're going for. If you need accuracy off more than 10%. I'm sorry. If accuracy upto 10% tolerance is allowed, then these changes are okay. But if you want the tolerances to be less than 1/3 of less than 10% of 21% then these changes. Uh, significant. So it depends on what you're going for. So if you want better accuracy, you can change bold meter and used a one meter, which has resistance off maybe a 1,000,000,000 makeup 1,000,000,000 or rather than when we go.

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