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Battery problem |
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| Feb19-07, 09:09 PM | #1 |
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Battery problem
1. The problem statement, all variables and given/known data
A schematic diagram incorporates a 12V battery in series with a 3.0 ohm resistor. The current then flows through 12 ohm and 4.0 ohm resistors in parallel with each other. a) calculate the current through and voltage across each resistor. 3.0ohm 4.0 ohm 12ohm 24ohm 4.8 ohm 4.0 ohm 2. Relevant equations V=IR I=V/R 3. The attempt at a solution Would I simply plug these into the equation and get an answer? 12/3=4A, 12/4=3A, 12/12=1A, 12/24=.5A, 12/4.8=2.5A 12/4=3A For voltage would I just plug these answers into the V=IR equation. |
| Feb19-07, 09:16 PM | #2 |
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You need to find the total resistance of the parallel pair of resistors, with that you can find the total circuit resistance. Once the total resistance is known you can find the total current. Given the current you can now compute the voltage drops. Given the voltage drop on the parallel network you can compute the individual resistance.
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| Feb19-07, 09:55 PM | #3 |
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I got the total resistance and it's 8. How do I calculate the current through each resistor not the total current.
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| Feb19-07, 10:09 PM | #4 |
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Battery problem
You could use V = IR. With your original equations:
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| Feb19-07, 10:14 PM | #5 |
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I got the total resistance to be 8. The total current to be 1.5 because 12/8=1.5, now how do I get the voltage drops?
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| Feb19-07, 10:47 PM | #6 |
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Well since you now have a circuit with one equivalent resistance in series with the voltage source and you know the total current, just work backwards. For example if your equivalent resistance contained the total resistance of R1+(R2||R3). We would work backwards by splitting up the equivalent resistance into two series resistors (with respect to the voltage source). One resistor would be R1 and the other would be R2||R3. Knowing the total current we can find the voltage drop across R1 and R2||R3 (R2 and R3 will have the same voltage drops becuase they are parallel).
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