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Current (I) = Voltage (V) / Resistance (R)
Man of Honour
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You can't do the first one (divide by zero), the rest seem fine.
These resistors were actually added in parallel? Then it makes sense, lower resistance = higher current. The battery voltage drops due to it's internal resistance, he would have observed the battery itself heating up with more resistors added.
It should be V=IR, but remember that resistors in parallel add as 1/RT = 1/R1 + 1/R2 etc. Check here:
http://www.1728.com/resistrs.htm
With 7 it's 1.4286ohm.
Hmm, it doesn't work out to those results :/
It should be V=IR, but remember that resistors in parallel add as 1/RT = 1/R1 + 1/R2 etc. Check here:
http://www.1728.com/resistrs.htm
With 7 it's 1.4286ohm.
Hmm, it doesn't work out to those results :/
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There's something very wrong with either the figures or the explaination of the experiment. The OP suggests that as the resistance in the circuit increases so does the current - with a fixed voltage source that's impossible.
Man of Honour
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Indeed. I remember noting all of the horrible flaws in my GCSE science experiments too
Ok firstly the units aren't mentioned.
The currents likely to be measured in milliamps not amps which might explain things.
Secondly 140Ohms in series with 0.778V gives about 5.5mA current which isn't far off what hes getting.
Theres also the internal resistance of the battery to add to that as well.
sid
The currents likely to be measured in milliamps not amps which might explain things.
Secondly 140Ohms in series with 0.778V gives about 5.5mA current which isn't far off what hes getting.
Theres also the internal resistance of the battery to add to that as well.
sid
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i don't suppose you could draw a circuit diagram?
Where was the voltmeter connected? Across the terminals of the battery? Across one or all of the resistors?
Where was the voltmeter connected? Across the terminals of the battery? Across one or all of the resistors?
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well i suppose you could always just take the gradient of the graph, and make an equation in the form y=mx+c. m is the gradient, x is the current, and c is the y intercept, so in this case, 1.6v. I used excel to work out the gradient, it should be -0.13.
therefore the voltage (y) is -0.13*current (x) + 1.6v.
EDIT: yep, this works well enough.
-0.13*3+1.6=1.21v, close enough to 1.18.
-0.13*6+1.6=0.82v. again, close enough.
So the final equation to calculate the voltage of the battery is y= -0.13*x+1.6
therefore the voltage (y) is -0.13*current (x) + 1.6v.
EDIT: yep, this works well enough.
-0.13*3+1.6=1.21v, close enough to 1.18.
-0.13*6+1.6=0.82v. again, close enough.
So the final equation to calculate the voltage of the battery is y= -0.13*x+1.6
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Associate
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No, it isn't too difficult, once it is explained how you do it, but they should have explained how to do this first, then asked them to do the experiment, and use what they have been told to find the equation.
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I would have thought so personally. I don't remember doing anything like this at school, we did occasionally wire up bulbs and simple things
