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Homework Help: Assigning voltage polarities

  1. Jan 9, 2009 #1
    I'm confused about how to assign the voltage polarities on circuit elements. Below is an example about what I'm confused about:

    http://img520.imageshack.us/img520/5075/circuitly2.jpg [Broken]

    So applying the Kirchoff's Law to the left loop, I get:

    -5 + 54000i1 - 1V + 186000i1 = 0

    So far so good. It's doing the Kirchoff's Law to the right loop that I get confused:

    If let's say we do a clockwise loop starting below the 8V, we'll get

    -186000i1 + v + ... + 8 = 0

    So in this case, how do I know where to place the positive terminal on the 1.8k resistor?

    because it does make a huge difference where I place it.

    Any help would be greatly appreciated, thank you.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 9, 2009 #2


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    Getting the polarities wrong is a common in KVL problems. You have to draw the current loops first. Pick any direction and draw one in each loop independent of the other loops, but once you have a chosen a current direction, pay close attention and stick to it. Then in the overlap branches the currents as you have identified them sum. That said, one can usually save some trouble by artful selection of the currents. In the right hand loop, if you choose clockwise current loop, lets call that i2, as you said (which is COUNTER to the given 30i1 current source), then the polarity is of the 1.8k is + on the left, - on the right for the clockwise current. The the voltage drop is 1.8i2. Note then for that choice i2 = MINUS 30i1.
  4. Jan 9, 2009 #3
    Thanks for your help mheslep. I'm still a bit confused though.

    So if I choose to use a clockwise loop for the left loop first, that'd make the middle resistor to have a plus polarity on top and negative polarity on the bottom.

    Next, if I choose to draw a clockwise loop for the right loop, does the polarities on the middle resistor which I've assigned while doing the left loop persist?

    This is the way the book did the example I have provided above.

    http://img404.imageshack.us/img404/998/circuit2ux0.jpg [Broken]

    So if I were to use the clockwise loop for the right loop, making the polarity of the 1.8k resister opposite of the book had provided, would I still get the same answer?

    Thanks again for any help. This whole assigning polarity is making my head hurt. The book barely talks about assigning polarity.
    Last edited by a moderator: May 3, 2017
  5. Jan 9, 2009 #4
    maybe the last circuit has too many elements for the concept I'm confused about. Here's another example:

    http://img111.imageshack.us/img111/7295/circuit3qm0.jpg [Broken]

    In this example, we'd first set the current i1 equal to the current exiting which is i2 and i3 at node a.

    After doing a clockwise loop on the left hand side, we have set the positive polarity on the 24ohm resistor at the top.

    Then if we were to take a clockwise loop for the right loop, we would assign the positive polarity to the top of the 8ohm resistor.

    That would make the equation, 8i3 - 24i2.

    But if we were to take a counter-clockwise loop for the right loop, we would assign the positive polarity to the bottom of the 8ohm resistor.

    That would make the equation 8i3 + 24i2.

    So the two equations don't match.
    Last edited by a moderator: May 3, 2017
  6. Jan 9, 2009 #5


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    You probably already know this, but whenever possible the first step is reduce the multiple parallel or series elements to an equivalent single element. In this case the parallel resistance of 24 and 8 ohms is 6 ohms. So the circuit reduces to 200=i1*(R+6).

    Regards, your comments, it appears you are somehow trying to mix KCL and KVL. Do one at a time. KVL: the sum of the voltages around a loop equals zero. See only one loop at a time. Define a single current that flows all the way around the loop. The way you defined i2 and i3 above they do not flow all the way around loop. Note that its valid to define a loop current because you can pull these loops apart via the principle of superposition, and then add them back together when you have an equation for each loop. Mechanically, you can 'cover up' all but the loop of interest and write down the voltages around the loop.

    Here's a rule set for KVL:

    Define the direction of charge moving clockwise around that righthand loop (i.e. a current, call it i2) then you trivially have either 8*i2 + 24*i2 = 0, or 8*-i2 + 24*-i2 = 0 if you like, showing that there's no use in summing voltages around a purely passive loop (i2=0).
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