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1. Hi all,

Thank you!
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2. (Original post by londoncricket)
Hi all,

Thank you!
ok so, first step:
the reading when switch is closed is 3 ohms for the resistance and 24v
2nd : as the switch is closed the resistance changes ( product over sum it is 2.25 ohms)
3rd: 3 / 2.25 is 0.75
4th:
0.75 x 24 = 18
5th: 24 - 18 = 6
3. Nevermind i messed up
4. (Original post by EggFriedRai)
ok so, first step:
the reading when switch is closed is 3 ohms for the resistance and 24v
2nd : as the switch is closed the resistance changes ( product over sum it is 2.25 ohms)
3rd: 3 / 2.25 is 0.75
4th:
0.75 x 24 = 18
5th: 24 - 18 = 6
Was the load connexted in parallel? I couldnt tell
5. (Original post by EggFriedRai)
ok so, first step:
the reading when switch is closed is 3 ohms for the resistance and 24v
2nd : as the switch is closed the resistance changes ( product over sum it is 2.25 ohms)
3rd: 3 / 2.25 is 0.75
4th:
0.75 x 24 = 18
5th: 24 - 18 = 6

What does "product over sum" mean?
6. (Original post by metrize)
Was the load connexted in parallel? I couldnt tell
you would assume it was set up reguarly
7. (Original post by EggFriedRai)
you would assume it was set up reguarly
I thought total resistance would be 12ohms
8. (Original post by londoncricket)

What does "product over sum" mean?
when you have two resistances in parallel, you do the product ( times them together ) nd then divide it by the sum ( add them up ) for the total resistance
9. (Original post by metrize)
I thought total resistance would be 12ohms
10. (Original post by EggFriedRai)
I couldnt tell from the question that it was in parallel so i assumed it was in series so i added the resistance
11. (Original post by metrize)
I couldnt tell from the question that it was in parallel so i assumed it was in series so i added the resistance
easily done mate dont worry
12. (Original post by EggFriedRai)
when you have two resistances in parallel, you do the product ( times them together ) nd then divide it by the sum ( add them up ) for the total resistance
Okay so this is the same as doing (1/A+1/B)^-1 if "A" and "B" were to be two resistances in parallel.

And so why would you calculate 0.75?
13. (Original post by londoncricket)
Okay so this is the same as doing (1/A+1/B)^-1 if "A" and "B" were to be two resistances in parallel.

And so why would you calculate 0.75?
to get the ratio to divide by

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