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Old 06-22-2015, 09:12 PM   #1
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Default Ohms Law Question! (School test review)

There are two questions I am a bit stumped on. I'm a bit confused on how to find the answers to them both. Although I know what the answers are because I have the review answer key, I still need help on knowing how to find the answers encase they show up on tomorrows test. Appreciate all your help!


Answer is A!



Answer is B!
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Old 06-22-2015, 09:20 PM   #2
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Quote:
Originally Posted by JasonCo View Post
There are two questions I am a bit stumped on. I'm a bit confused on how to find the answers to them both. Although I know what the answers are because I have the review answer key, I still need help on knowing how to find the answers encase they show up on tomorrows test. Appreciate all your help!


Answer is A!



Answer is B!
Regarding question 19:
Resistance appliances (incandescent light bulbs, electric heaters, etc) are rated in watts, but this rating applies specifically to that appliance's voltage rating, if the appliance is used at a different voltage, it will actually dissipate a different amount of energy. Think of this concept as you try to work out this question.

Regarding question 22:
There is insufficient information to answer this question.
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Old 06-22-2015, 10:00 PM   #3
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#19 - as above - you need to figure out resistance of the lamps which is a fixed value regardless of the voltage, current or power.

#22 - So basic question of what happens when MWBC loses its neutral - you ignore other loads in the panel. It gives you a simple series circuit with load 1 and load 2. Just figure out the resistance of each of the two loads individually based on 120v. Then calculate their voltage drops across a 240v source.

Last edited by Stuff; 06-22-2015 at 10:05 PM.
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Old 06-22-2015, 10:09 PM   #4
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#19 - as above - you need to figure out resistance of the lamps which is a fixed value regardless of the voltage, current or power.

#22 - So basic question of what happens when MWBC loses its neutral - you ignore other loads in the panel. It gives you a simple series circuit with load 1 and load 2. Just figure out the resistance of each of the two loads individually based on 120v. Then calculate their voltage drops across a 240v source.
Yes that is true, but the question doesn't tell you what the loads plugged into the two halves of the split receptacle are, so there is insufficient information to answer the question.
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Old 06-22-2015, 10:26 PM   #5
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Yes that is true, but the question doesn't tell you what the loads plugged into the two halves of the split receptacle are, so there is insufficient information to answer the question.
Since this is a test question and not the real world I assume that the nameplate info given is accurate and 100% resistive for these calculations.
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Old 06-22-2015, 10:31 PM   #6
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My mistake, I missed some of the writing. All the required info is indeed there.
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Old 06-22-2015, 10:50 PM   #7
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Thanks for the help! I've taken your advice on problem 19 and have been trying to solve it. Let me update this with what I have done, one sec
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Old 06-22-2015, 11:04 PM   #8
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Okay so the resistance of Circuit 20 would be (130^2) / (150w) = 112.666
and there is 5 luminaires so I guess I am assuming they are in parallel? So the parallel resistance formula would be 1 / [(1/112.666) + (1/112.666) + (1/112.666) + (1/112.666) + (1/112.666)] = 22.533 = is the resistance of circuit 20.

Now circuit 22 has 3 loads.
First load: 130 / 5.6 = 23.21 Resistance
Second Load: 130 / 1.8 = 72.22 resistance
Third load: (130^2) / 550 = 30.73 resistance

And I am to assume they are in parallel? So the parallel equation for resistance would be 1 / [(1/23.21) + (1/72.22) + (1/30.73)] = 11.18 resistance for circuit 22.

Now that I have both resistances, not even sure if I did that right? But having both, I just add them together? 11.18 + 22.53 = 33.71 resistance for both 20 and 22 circuits.

Idk I think I am doing something wrong? And also I'm unsure what to do after these steps
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Old 06-22-2015, 11:14 PM   #9
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Okay so the resistance of Circuit 20 would be (130^2) / (150w) = 112.666
and there is 5 luminaires so I guess I am assuming they are in parallel? So the parallel resistance formula would be 1 / [(1/112.666) + (1/112.666) + (1/112.666) + (1/112.666) + (1/112.666)] = 22.533 = is the resistance of circuit 20.

Now circuit 22 has 3 loads.
First load: 130 / 5.6 = 23.21 Resistance
Second Load: 130 / 1.8 = 72.22 resistance
Third load: (130^2) / 550 = 30.73 resistance

And I am to assume they are in parallel? So the parallel equation for resistance would be 1 / [(1/23.21) + (1/72.22) + (1/30.73)] = 11.18 resistance for circuit 22.

Now that I have both resistances, not even sure if I did that right? But having both, I just add them together? 11.18 + 22.53 = 33.71 resistance for both 20 and 22 circuits.

Idk I think I am doing something wrong? And also I'm unsure what to do after these steps
Couple problems:
Circuit 22 is a red herring. Ignore it. The question asks you about the CURRENT drawn by circuit 20. So you do need to find resistance, but only in the pursuit of the circuit current. As for your resistance calculations for circuit 20, the formulas are correct, but go back and double check just how many lamps you are dealing with, I think you'll find you've made an error. After you figure that out, finding the current will be a breeze.
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Old 06-25-2015, 12:54 AM   #10
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You will never learn to be a good electrician untill you learn that you must scribble in straight lines. Do not cover over your questions and anwsers in spaghetti. It made me cry a little as I looked at your pic.
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Old 06-25-2015, 01:11 AM   #11
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Read this

http://www.ecmweb.com/content/multiw...n-be-dangerous
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