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Voltage distrubution in a circuit

  1. Apr 25, 2012 #1
    Hi guys, I came across a pretty weird question regarding electricity in a parallel circuit. There are two branches of the circuit with a resistor and a light dependant resistor on each branch. The weird part is there a wire is connected in between the resistor and LDR for both branches which is then connected to a logic box. So the logic box has 2 input wires but only one output wire which is also connected to a bulb, lastly it is connected to the neutral wire leading back to the power source. So I'm not sure how the voltage will be distributed amongst the two resistor, LDRs, logic box and bulb. I'm not sure how will the voltage will be divided once the current in the two branches meet the wire with the logic gate. Thanks for the help guys! :smile:
  2. jcsd
  3. Apr 25, 2012 #2
    I hope I managed to follow you. Maybe a link with the circuit would help. If I understand correctly than you're asking why there's another resistor connected to the LDR.
    The reason the LDR is connected in series with a resistor is to limit the current that reaches the input.
    The distibution of voltage depends on the situation and the LDR's resistance at the time. The voltage in parallel is equal and the sum of the currents in parallel should equal the initial current. Without the parameters you can't predict the initial voltage distribution.
  4. Apr 25, 2012 #3
    Hi michelle15g, thanks for the reply i have drawn out the picture and my further questions are typed out on it too to better convey the message. Do let me know if i was unclear on any parts of the questions.

    diagram: http://postimage.org/image/xg3p2s8xn/
    (see the picture before reading this)
    That being said, what will happen if i connect two cells in parallel but one cell has 10V and the other 5V? What will happen if something like this happens?

    Thanks for the help! :smile:
  5. Apr 25, 2012 #4
    Hi sgstudent,
    The voltage on a resistor or LDR can change. In other words - the distribution doesn't have to be the same for the resistors in parallel. If you add an identical resistor - it won't have the same voltage distribution as the others.
    The current through the resistors doesn't have to be the same. Their sum must equal the original current.
    The logic box can decide what voltage it requires for a logical True or false. The cutoff is usually 1V, 3V or 5V.
    The logic box would decide what combination of True and Falses will light the bulb.
    It's kind of hard to imagine a state where one LDR would receive more light therefore altering the resistance ratio with the resistor but it exists. Some appliances use this fact to shift the angle. The difference in ratio would result in a difference of logical inputs.
    Lets look at the case where one LDR receives an extreme amount of light while the other none.
    Its resistance would turn to a small fraction of the orginal resistance. The series' resistance would change to R + frac of LDR. Most of the current would choose to pass through this path.
    The potential at the end of the path would be a logical one.
    There are a few circuit simulation sites that could be useful. I've tried a few and I think they will help you out. Just try out a few combinations and see what happens.
    Good Luck :)
  6. Apr 25, 2012 #5
    oh, but for simplicity sake how will the voltage be split amongst the loads if the resistors and the LDRs both had the same resistance. Also, since the logic box is quite complicated then if we ignore it and just have the bulb connected will what it work like the assumptions i gave?

    thanks for the help! :smile:
  7. Apr 25, 2012 #6
    Hi Michelle, I asked my sister and she said that I can use the trace method where I follow through on the different possible branches and the 6V will be accordingly distributed. Will this work out? Will it enable the logic box (now replaced with the bulb) to have the same voltage for all possible branch? Thanks for the help!
  8. Apr 26, 2012 #7


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    Staff: Mentor

    It's equivalent to connecting the two in series. The 10V battery will force current through the 5V battery. If these are dry cells and you connect them for more than a brief moment, the 5V may be damaged or explode, as these cells are not designed to be charged. In addition, the 10V battery is probably not designed to supply the high current that this action is likely to cause, and it may overheat or have its life shortened.

    It is not possible to say what the voltage measured at the terminals of the batteries would be, as the calculations require details of the electrical characteristic of the cells when experiencing high currents and in the directions as would arise here. Though we can be sure 5 < VT<10V
  9. Apr 26, 2012 #8
    Hi NascentOxygen. Thanks for the info it really helped. But what about the voltage in the logic box? How do I find out that voltage value? Because I've only been exposed to parallel circuit connected to one single branch so now when 2 wires connected to two different branches and connected to a single appliance, I don't know how to determine voltage through it. In normal ones, the voltage is the same because Pd=potential at final-potential at initial. At the two points I can determine the potential of the points but I brave no idea how to go about solving it. My guess would be that from one of the main branches, i get the voltage at that point and at the neutral, but there are 2 of them so I'm pretty sure that's not right. As for the current and resistance, I'm completely clueless about it. Thanks for the help!
    Last edited: Apr 27, 2012
  10. Apr 27, 2012 #9
    I think that when you connect to a logic box - you don't care what voltage you get exactly. What you care about is getting a logical one or logical zero. BTW, what kind of bulb is connected to the logic box? Is there a separate VCC connected to it allowing it to turn on the light? I think the logic box supplies a different VCC for the bulb because without an amplifier, it would be difficult to light the bulb.
  11. Apr 27, 2012 #10
    I was just wondering how will the voltage be distributed if there was no logic box, just a bulb. Cos I'm not sure how to determine voltage even if the logic box wasn't there as the wires connection is really confusing to me. That being said I'm also not sure how to determine current or resistance of the circuit. Could you teach me on how to determine them? Thanks for the help! :smile:
  12. Apr 27, 2012 #11
    Yeah, I understood what you meant but I think there's more to the logic box than logic. A simple LED might turn it into a simpler circuit.
    This site explains a bit about circuit in series and in parallel.
    After you read that - you should read a bit about changeable resistance.
  13. Apr 27, 2012 #12
    Hi Michelle! :smile: I understand the formulas relating them together but I don't know how to apply them into this circuit because of its weird nature where two wires are connected to 2 separate parallel branches. I can get the potential at the 2 points but there are 2 points so I'm not sure which one to use or both of them since at the other end the potential is 0V. Also for the resistance, do I have to get the total current through it and divide by the 6V? But that being said I still have to find the voltage across the bulb... Thanks for the help!
    Last edited: Apr 27, 2012
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