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Homework Help: Archived Potential Divider

  1. Jan 16, 2005 #1
    Hi, I'm having a bit of trouble with a potential divider question. I have an answer but I'm not sure if its right and if not, why its not right. I've attached a crude drawing of my problem below, I have a question asking what the output voltage is (between P and Q terminals) with the given resistances in ohms and battery of p.d. 6V. I think what is confusing me is the line drawn between C and D. Anyway, I know that the output voltage is the resistance across where the circuit is divided x current in circuit. The problem I had was whether to work out the effective resistance of the 2 resistors in parallel (the 800 and 400). I worked out total resistance in the whole circuit to be 400ohms (1/600 + 1/1200=1/Rtot). The current turned out (V=IRtot) as 0.015A. So working out output voltage I got 0.015 x Reff (1/400 + 1/800= 1/Reff) which gave an output of 4V which seemed like a reasonable answer. This was by simplifying the 4 resistors into 2 resistors in series. The problem with this is that the next question asks if there is a current flowing in this wire (CD) and if so what direction (as conventional current). Given how I have simplified the 4 resistors, and that the P output terminal will be +ve I would say that conventional current will be flowing from C to D? But I am confused because there are actually 4 resistors which is the problem.

    Sorry I've made such an essay out of it, but I'm very confused, I've had no trouble with the more simple potential divider questions but this 1 was totally different. I'm not entirely convinced of my answer to the output voltage. I would appreciate if someone did a quick check on my work and maybe explain this circuit to me and anything I've got wrong. Thanks very much, Jamie

    Attached Files:

  2. jcsd
  3. Feb 7, 2016 #2


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    Staff: Mentor

    Thanks to the connection between C and D both the top two resistors and the bottom two resistors form parallel pairs.


    Find the two equivalent resistances and the circuit is reduced to the classic voltage divider scenario.

    The top resistors reduce to an equivalent of 400/3 Ω. The bottom pair reduce to 800/3 Ω. So they are in a 1 to 2 ratio. That makes the output of the divider 2/3 of the input (that is, 2/(2 + 1) = 2/3).

    So the output across terminal PQ will be 4 V.
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