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Tricky(?) resistance question in a series circuit |
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| Jul24-07, 04:24 PM | #1 |
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Tricky(?) resistance question in a series circuit
1. The problem statement, all variables and given/known data
Two resistances, R1 and R2, are connected in series with a 12V battery. When R2 is removed (leaving R1 only), the current increases by 0.450 A. When R1 is removed (leaving R2 only), the current increases by 0.225 A. What are the two resistances? 2. Relevant equations [tex]V = IR[/tex] 3. The attempt at a solution I've tried to solve this algebraically, but I keep ending at a point with three unknowns and no realistic way to solve them. The fact that the current bumps are at a 1:2 ratio suggests that something similar should happen with the resistors, but hell if I know where to start. |
| Jul24-07, 04:43 PM | #2 |
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| Jul24-07, 05:49 PM | #3 |
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So from that, in scenario #1, V = (I+0.450)(R-R2), which factors out to: 12 = IR - IR2 + 0.450R - 0.450R2 Scenario #2, V = (I+0.225)(R-R1), or: 12 = IR - IR1 + 0.225R - 0.225R1. With three unknowns in each equation, it just feels like I'm going about this the wrong way. |
| Jul24-07, 06:05 PM | #4 |
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Mentor
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Tricky(?) resistance question in a series circuit
Try focusing on the main parts:
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| Jul24-07, 06:16 PM | #5 |
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It seems like berkeman has already helped you, but I did say I would help and I am a man of my word!
So, even if I am just echoing berkeman, here goes.You have two equations and three unknowns. It looks to me like what you need is another equation, or a way to get rid of one of the unknowns. I'm going to go for the former. Here's my hint: What about the original circuit with two resistors? Can you analyze this to get more information? |
| Jul24-07, 06:24 PM | #6 |
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hrm... I'm really not catching on to the best way to simplify this a bit more (V = IR? 12 = I(R1+R2)? etc...) Of an entire assignment of similar questions, this is the only one I'm missing.
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| Jul24-07, 06:44 PM | #7 |
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You now have: 12=I(R1+R2) 12 = IR - IR2 + 0.450R - 0.450R2 12 = IR - IR1 + 0.225R - 0.225R1 Here are 3 equations with 3 unknowns. You should be able to solve for all three now. |
| Jul25-07, 10:17 AM | #8 |
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Gotcha, thanks.
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