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The current through an individual resistor

  1. Oct 21, 2014 #1
    The following circuit has been modelled using some modelling software (Micro-Cap 11). upload_2014-10-21_19-51-33.png

    I was wondering why current is only shown at the source and not in R1, R_Unknown and R3.
    I know current is meant to be the same in series although I am confused by the effects of the resistors.
    (This is not a homework question, just pure curiosity.)
     
  2. jcsd
  3. Oct 21, 2014 #2
    Why are you confused? Seems like you already know the answer.
     
  4. Oct 21, 2014 #3
    Yes, the current is the same everywhere in this circuit, so it makes sense to show it only once.

    For the other numbers shown, it seems to be like this:

    R_total = R1 + R_unknown + R3
    U = 6V
    I = 1.855 mA

    Voltage drop over R1, measured across R1: U_R1 = R1 * I = 1.855 V
    So, measuring the voltage after R1 to ground would give you U - U_R1 = 4.145 V
    The voltage drop of R3 is the same, so also 1.855 V.

    And for R_unknown it should be:
    U_R_unknown = U - U_R1 - U_R3 = 2.29 V
    (Since all voltage drops together should sum up to the total voltage of the battery... Kirchhoff's Voltage Law.)

    R_total = U / I = ~3234.5 Ohm

    U = R_total * I
    U = ( R1 + R_unknown + R3 ) * I

    Solving this for R_unknown:
    R_unknown = U / I - R1 - R3 = R_total - R1 - R3 = ~1234.5 Ohm

    The voltage drop of R_unknown seems to be right:
    U_R_unknown = R_unknown * I = 2.29 V

    The sum of all voltage drops in series should give you the total voltage again (check 2.29 V + 2 * 1.855 V = 6 V).

    So the number shown after R_unknown (1.855 V with the pink frame) should be the voltage measured across R3...

    Currently trying to figure out all this, as well... ;)
     
    Last edited: Oct 21, 2014
  5. Oct 21, 2014 #4
    So the current in R_Unknown will be the same. I would have thought the resistor would of restricted the flow of current so in the resistor current would be less?
     
  6. Oct 21, 2014 #5
    No, the current is same in all resistors in serises. This follows from the law of conservation of electric charges.
     
  7. Oct 21, 2014 #6
    Ok thank you. So why do voltage drops change then.
     
  8. Oct 21, 2014 #7
    Ohm's law?
     
  9. Oct 22, 2014 #8

    sophiecentaur

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    The voltage drop is due to the energy dissipated as the charges pass through the component. All such circuits settle down into the condition that the Power dissipated in all the resistors (given by IR) is equal to the Power input from the Source. That's the Conservation of Energy thing (Kirchoff's Second Law). The Charge conservation law is his first law.

    I suggest you Google Kirchoff's Laws and read all about them. There is bound to be something at an appropriate level which will allow you to grasp what's going on. Remember - nothing 'knows what's going on' in a circuit; every element in the circuit just responds to the conditions immediately around it - so the 'sharing out' of Voltage is something that actually takes time and that time is governed, basically, by the speed of the initial electrical impulse round the circuit as the switch is turned on (a bit less than the speed of light).
     
  10. Oct 22, 2014 #9

    sophiecentaur

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    OHM's law only applies to metals at a constant temperature. It is only a sub-set of the way V and I relate in circuits. It's not one of those fundamental laws; rather, it's a description of behaviour of some things under some conditions.
     
  11. Oct 22, 2014 #10
    I didn't notice OP mentioned somewhere nonlinear resistances.
     
  12. Oct 22, 2014 #11

    sophiecentaur

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    True. But why use a term that applies in only some circumstances? The ' reason' for what he was asking about can be explained using Kirchoffs Laws, which do not assume Ohm's Law behaviour and conditions.
    Fact is that "R = V/I" is not Ohm's law - look it up and see that Ohm's Law includes the "metal" bit and the "temperature" bit.
     
  13. Oct 22, 2014 #12
    Why the Ohms Law bashing ?- In circuit theory V=I*R ( Voltage is proportional to current and resistance) applies to 99.99+ % - I'll even argue about the constant temperature issue, as there is no time element in this formula is is a static calculation and there is nothing stating that R must be constant. As for the metals requirement that also is not too accurate, but Ohmic materials- the OPs case is basic circuit theory and discouraging the application of Ohms law is counterproductive.
    ... just like Newton's Law of Gravitation is technically invalid - it is taught as the basic principal and the analysis derived from it is considered valid in all but exceptional cases. Why bring up special relativity(speed of light - really?) to discuss how an apple falls from the tree.

    IMO - If Ohms law is good enough for an MIT education -- it is fine here.
     
  14. Oct 22, 2014 #13

    nsaspook

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    I don't think anyone is bashing Ohm's Law but the world of electrical phenomena is in general not linear in the large signal world. Even a small incandescent lamp has "nonlinearity" of resistance that varies with voltage. Ohm's Law should be seen as a definition of resistance for a given a voltage and current that might be linear in a range of values. Ohm's Law is the exception once you get past simple textbook circuits of ideal resistors and perfect voltage sources as Ohm's law requires proportionality between current and voltage.
     
    Last edited: Oct 22, 2014
  15. Oct 22, 2014 #14

    sophiecentaur

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    Does Ohm's law apply to a diode, a Field effect transistor, a thermistor, a Capacitor or an Inductor? If you assume that the resistance of a thermistor is constant and independent of it's place in a circuit, you will soon come unstuck. You can't assign an "Ohm's Law" resistance value to one of them but you can calculate the V or I under specific conditions. I think you are misinterpreting my "bashing" exercise. I am bashing the over liberal use of the expression "Ohm's Law" as if it's some sort of explanation, rather than a description of behaviour.
    I really cannot believe that you were ever told that R - V/I is all there is to Ohm's Law in MIT. (When I was at secondary School, they served us better than that). How have I "discouraged the use of "Ohm's Law? Au contraire, I am encouraging the use of Ohm's Law where it actually applies and the use of the correct terminology -i.e. V = IR everywhere. I don't understand this defence of merely sloppy terminology. There is no equivalent sloppiness when dealing with Impedances in a circuit so why do it with pure Resistors?

    Newton's 'Law of Gravitation' works (really works very well') for all objects under the same broad but finite set of gravitational conditions. It's an entirely different matter from a description of the behaviour of some particular devices.
     
  16. Oct 22, 2014 #15
    We are lucky that OP didn't put AC source in his circuit. Someone might complain that resistors aren't purelly resistive but have parasitic inductance and capacitance and model isn't adequateo0)
     
  17. Oct 22, 2014 #16

    sophiecentaur

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    You have obviously missed my point. I am not talking about the details of measurement and characterising components appropriately. But would you really use the terms "Ohm's Law" to describe the action of the Resistive part of an Impedance and then use some other term for the Reactive part?
    The reason I picked up on this, initially was that the term "Ohm's Law" was used, on it's own, in one of your posts, as if that explained everything. It doesn't. The ratio of volts over current is not "Ohm's Law", although we often find it doesn't change with current. Aren't we (PF) trying to help people with this subject and to encourage a bit of care and rigour in their arguments?
     
  18. Oct 22, 2014 #17
    Judging by his questions I think cavalieri doesn't know basic things about simple DC circuits with linear elements. First things first. You don't learn to run before you learn to crawl.
     
  19. Oct 22, 2014 #18

    sophiecentaur

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    So why tell him that R = V/I is Ohm's Law? Just tell him the Resistance is V/I, which is "crawling" in the right direction. For those things with two wires on them and coloured bands around them, the ratio is pretty much the same until they burn out.
    I still think you are missing my point entirely on this one.
    Perhaps you should actually read about Ohm's Law and not just rely on your memory. Even the Wiki page on it is constantly pointing out the failure of many components to follow Ohm's Law - because it's just an empirical observation. It is not universal and not an 'explanation' of anything.
     
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