Resistors with Terminals (if Possible)

Discussion in 'General Electronics Chat' started by bluetooth tamer, Jan 16, 2015.

  1. bluetooth tamer

    Thread Starter New Member

    Jan 16, 2015
    I have a very basic question about reading resistors with terminals. I got this diagram from my notes from lecture during Fall. I'm trying to do some review before my Spring Circuit Analysis class begins.

    Basically I would like to know if both my diagrams are valid and if so what does the difference in the polarity of terminals translate to?

    Equation: P = I^2 R

    In the first diagram I have the resistor absorbs power equal to 40 watts.

    Because the next diagram has the polarity flipped, would it mean that the resistor delivers power equal to 40 watts?
  2. kubeek


    Sep 20, 2005
    No, that would mean that either the current in second picture is acutally -2A, or that you switched from conventional current flow to electron current flow, because we all know resistors don´t generate power.
    If it were not a resistor but some two terminal black box, then your previous statement would be true, in first picture it is absorbing power and in the second it is generating power (from the conventional current flow point of view).
  3. ian field

    AAC Fanatic!

    Oct 27, 2012
    Resistors convert electrical energy into heat energy - unfortunately, they can't reverse that process.
  4. WBahn


    Mar 31, 2012
    It depends on what the voltage polarity symbols in your diagram is meant to convey. Often it is merely to establish the "symbolic voltage" across the resistor. This just means that if I later tell you that the voltage across the resistor is +10V, you know that the terminal with the + sign is 10V greater than the terminal with the - sign. Similarly, if I tell you that the voltage across the resistor is -10V, you know that the terminal with the + sign is 10V less than the terminal with the minus sign. This is because the symbol voltage, Vr, across the resistor is defined to be

    Vr = (V+) - (V-)

    where V+ is the voltage at the + terminal and V- is the voltage on the minus terminal.

    Since this is a resistor which can only absorb electrical energy, what the diagrams then tell us is that the voltage across the resistor in the first case is positive and that it is negative in the second case.
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  5. bluetooth tamer

    Thread Starter New Member

    Jan 16, 2015
    I haven't had a chance to look over my lecture notes until this morning. Thank you guys. It seems like I got two different answers, #4 being more hypothetical. I would like to show a more concrete example with varying terminals from my notes.

    I have it written down that when current is entering a negative terminal of a resistor the correct equation to use to find voltage is V = -IR. However, I can't find this equation online or in my circuits book.

    If at any point you see a mistake please do point it out. The diagram has Vs, V1, V2, and V3 as the unknowns but because current and resistance is given. I solved this as follows:

    ΣV = 0 in a closed loop
    -Vs + V1 + V2 - V3 = 0 Eq.1
    V1 = 10A * 1Ω
    V2 = 10A * 2Ω
    V3 = -IR = -10A * 3Ω
    Plug it into Eq.1:

    -Vs + 10V +20V - (-30V) = 0
    Vs = 60V​

    My questions are (a) is V = -IR valid/correctly implemented and (b) if so why would the terminals on a resistor affect whether negative or positive voltage differences occur? (Doesn't a resistor always have a voltage drop?)
  6. WBahn


    Mar 31, 2012
    Without looking over your work, I'll answer your (b) question. Imagine a resistor in front of you oriented left to right. It has current flowing in it (note that I haven't said what direction the current is flowing). It has a voltage drop across it, but which terminal has the higher voltage and which terminal has the lower voltage? You don't know, because that depends not only on how much current is flowing, but also on which direction it is flowing, right? So both magnitude and direction matter and you haven't properly indicated either a current or a voltage unless you indicate both magnitude and polarity.

    So now I tell you that the voltage across this 1Ω resistor is 10V. If I ask you what the current is that is flowing in the resistor, you can only tell me the magnitude of the current but that's not enough because you also need to be able to tell me the direction that it is flowing in, because that matters.

    Now I tell you that the voltage across the resistor is defined such that Vr is positive at the left terminal relative to the right terminal or, more specifically, that Vr = V(left) - V(right). This is the "symbolic voltage" and merely establishes a reference for us. The actual value of Vr can be positive, negative, or zero. If I now tell you that the voltage across the resistor is 10V, you know since Vr>0 that the voltage on the left is greater than the voltage on the right and therefore you have 10A of current going from left to right.

    Had I defined the voltage across the resistor to be positive at the right terminal, then the voltage I would have given you for Vr would have been -10V. In either case the voltage of the left terminal is higher than the voltage at the right terminal and current is flowing from left to right.

    But talking about "current flowing left to right" in order to indicate direction is cumbersome and leads to all kinds of confusion quickly. So we define a "symbolic current" through the resistor, again to use only as a reference. If I define the symbolic current, Ir, is positive from left to right, then you would conclude that the current in this resistor is +10A. But if I had defined the symbolic current to be from right to left, you would have concluded that the current is -10A. In either case, the current is flowing left to right.

    So I need to define my symbolic voltage and symbolic current for a given device and I can arbitrarily pick the polarity of each and get correct answers, but to help minimize math errors we widely use what is known as the "passive sign convention" which is that we define the direction of the symbolic current such that it enters the positive terminal (as indicated by the symbolic voltage) of a passive device such as a resistor (or any device we choose to assume is a load). We do the opposite for any device we choose to assume is a source.
  7. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    Let’s answer backwards:

    Resistors with no current thru them have no drop. All others do have a voltage drop.

    The terminals matter when determining the voltage. Put a voltmeter across a resistor, note the voltage. Now swap the leads and measure it again. You get the same voltage but the opposite polarity (or sign).

    Your analysis using Kirchhoff's voltage law (KVL) is correct and complete. You made an arbitrary choice as to the polarity of V3 and the math lead you to the correct answer.

    Isn’t that nice? ;-)
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