# issac physics helpWatch

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#1
A simple series circuit is made up of two resistors of resistances 4.0Ω and 8.0Ω and two cells of 2.0V and 3.0V. Unfortunately one of the cells is connected the wrong way round and the current is rather small. The cell is then reconnected in the circuit the correct way round.

By what factor does the current that flows through the 4.0Ω

i got 5??
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3 weeks ago
#2
Any chance you could send the question or a picture of it?
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3 weeks ago
#3
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3 weeks ago
#4
Given what you’ve said above, and assuming that I’ve interpreted it correctly, you should find that the voltage across the 4 ohm resistor in the first setup is 1/3 and 5/3 in the second. As such, the current in the first setup is 1/12 and in the other it’s 5/12. This means it’s increased by a factor of 5. Is 5 not the right answer?
Last edited by hthain14; 3 weeks ago
0
#5
(Original post by hthain14)
Given what you’ve said above, and assuming that I’ve interpreted it correctly, you should find that the voltage across the 4 ohm resistor in the first setup is 1/3 and 5/3 in the second. As such, the current in the first setup is 1/12 and in the other it’s 5/12. This means it’s increased by a factor of 5. Is 5 not the right answer?
i solved it.

it was 5.0
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