Proving Ohm's Law between 2 circuits

AI Thread Summary
The discussion revolves around a physics lab assignment to prove Ohm's Law using two different circuit configurations. Circuit A measures total current but does not isolate the current through the resistor, while Circuit B measures voltage across a combination of resistors rather than just the resistor of interest. The results from both circuits differ, leading to confusion about the discrepancies. Participants suggest that the differences may stem from the finite resistance of the measuring instruments, particularly if the equipment is poorly maintained. The issue highlights the importance of accurate measurements in experimental physics to validate Ohm's Law.
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Homework Statement


Hello guys. I've ran into this conceptual problem. I was given into physic's lab to prove Ohm's law by mounting a circuit in two ways, and calculate both Voltage and Intensity.


The first circuit, the A (Please refer to Circuit A.jpg), has the inconvenient in which the ammeter does not measure the current (I) which passes through the resistance *(R), but the total intensity (I) of the circuit.

The second one, the B (Please refer to Circuit B.jpg), has the inconvenient in which the voltmeter does not really measure the potential difference between the ends of the resistance, but between the ends of the association of resistance in series with the ammeter:


They should give me the same. Problem is, the results between circuit A and circuit B gave me different, and I must know why. The teacher said something about the resistance, but it didn't quite grasp what he said. I'm recurring here for help, any ideas?


Thanks a lot in advance.

P.S.: I've left you the results from Circuit A and B. They both were done by making the Voltage as the independent variable.



Homework Equations



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The Attempt at a Solution


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Attachments

  • Circuit A.jpg
    Circuit A.jpg
    7.5 KB · Views: 492
  • Circuit B.jpg
    Circuit B.jpg
    7.1 KB · Views: 506
  • Circuit A and BResults.PNG
    Circuit A and BResults.PNG
    12 KB · Views: 541
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what are your units for current? A, mA, uA, nA?

most of the time I would say that has something to do with the finite resistance values of you voltmeter and ammeter, but if your current values are measured in whole amperes then you probably have other problems as well, or your equipments are particularly lousy...
 
Thanks for the quick reply.

Sorry for not posting the units! But yest, they are measured in Ampere (A).

Being the lousy equipment could be a very big possibility, since I don't believe they are taken good care of them.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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