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Question about circuits

  1. Nov 11, 2006 #1
    I'm having an obvious conceptual problem with circuit theory.

    Suppose I have a source of emf E connected in series with a resistor R. The loop rule suggests that the current through the resistor is i = V/R.

    Now consider a parallel system of two sources of emf, connected in series with a resistor. Application of the loop rule to a loop through either source suggests that the current through the resistor is i = V/R. If I now replace one source with low resistance wire, then very little current will pass through the resistor, and yet the loop rule still suggests that the current through the the resistor is i = V/R?

    What am I missing here?

    Thanks

    James
     
  2. jcsd
  3. Nov 11, 2006 #2

    OlderDan

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    The current through the resistor is I = V/R in all the cases you mentioned. A low resistance path parallel to the resistor can only change the current through that resistor if it affects the voltage across the resistor. Such a low resistance path is called a shunt. Any real source of emf like a bettery has internal resistance that we often ingore in simple circuit theory. The shunt causes a huge increse in the current through the battery, which lowers its terminal voltage, which reduces the current through the resistor. In an ideal battery (no intermal resistance) the current is umlimited and a huge current will flow through the shunt with no effect on the current through the resistor.
     
  4. Nov 11, 2006 #3
    Hi Dan,
    So if we consider the single source as an ideal battery with a small resistance r, then integrating around the loop containing the shunt and the battery gives
    [itex]\mathcal{E} - ir = 0[/itex]. which gives a large current as you suggest. On the other hand, if we integrate around the loop containing the battery and the resistor we get [itex]\mathcal{E} - i(r+R) = 0[/itex] gives a totally different current. Is there any way to mathematically resolve this contradiction?

    Thanks

    James
     
  5. Nov 11, 2006 #4

    nrqed

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    Hi James. There is no contradiction. The current will be *different* if the battery is connected to an ideal wire or connected to a resistor R. Notice that the current "i" in your two equations is NOT the same current. Maybe you think that there is a contradiction because you are assuming that they should be equal? They are not.

    A better notation would be to call the currents [itex] i_1 [/itex] and [itex] i_2 [/itex] where the index 1 and 2 would refer to the circuit used. As soon as you change the circuit, the current changes.

    Hope this makes sense.
     
  6. Nov 11, 2006 #5

    nrqed

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    Another way to see this is to be more careful with [itex] V = R i [/itex]. A better notation is
    [tex] (\Delta V)_{\rm across~R} = R i_{\rm across~R} [/tex]
    Every time you use this equation, you must ask yourself which resistance you are using and then you must use the potential difference across that resistance.
     
  7. Nov 11, 2006 #6
    Hi nrqed,

    No, I think you have misunderstood the thought experiment. The experiment consists of a battery with internal resistance r in parallel with a zero-resistance wire connected to a resistor R. There does indeed appear to be a contradiction when one attempts to integrate around the loop containing the shunt and battery or the loop containing the battery and the resistor as I pointed out before. In both cases the current through the resistor should be the same value because we are talking about the _same circuit_.

    James
     
  8. Nov 13, 2006 #7

    OlderDan

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    Let's go back to your original problem with 2 sources of emf in parallel feeding a single resistor. Assume both sources are identical with emf = E and internal resistance = r. Assume there is a left loop and a right loop with the resistor R in the middle. The currents are I1 for the left source I2 for the right source, and I for the resistor in the middle. The loop equations and current equation are

    E = I1*r + I*R
    E = I2*r + I*R
    I = I1 + I2

    The solution is

    I1 = I2 = I/2
    I = E/(R + r/2)

    If you replace the second source with a shunt of resistance r these become

    E = I1*r + I*R
    0 = I2*r + I*R
    I = I1 + I2

    I'll leave it to you to work out the algebra, but the solution for I is definitely not the same as in the first case.

    If the second source is replaced with a short, the solution is again simple

    E = I1*r + I*R
    0 = I2*0 + I*R
    I = I1 + I2

    I1 = -I2 = E/r
    I = 0
     
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