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Electronics - calculating voltages

  1. Jan 19, 2013 #1
    1. The problem statement, all variables and given/known data
    See the picture, problem 12:

    2. Relevant equations
    Ohm's Law, V = IR, et al

    3. The attempt at a solution
    So to be clear, I am being asked to calculate the voltage AT a point, not the voltage DROP across any point.
    It SEEMS easy... My initial thoughts are:
    A: V = 0
    B: V = 0
    C: V = 6

    The reason: Voltage drop from one side of the battery to the other should be 6v, therefore voltage at A should be 0. All voltage should have been removed by resistance at this point.

    At point C, the voltage has just left the battery, has not met any resistance, and therefore is still 6v. Pretty simple.

    At point B, this is connected to the ground, therefore voltage is at 0v (am I right?)

    My question is:
    When you connect a ground like this in the middle, could you say that there's a negative voltage drop (positive voltage gain) from point B to point A?

    Moreover, will that positive voltage gain have to be equivalent to the voltage drop across the 10kΩ resistor?

    Or am I thinking about this all wrong? I need to figure this part out before I can calculate current across the 10kΩ resistor...
  2. jcsd
  3. Jan 19, 2013 #2


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    Your answer is not correct.

    Voltages are allways relative to somewhere. Try using point B (earth) at your voltage reference (eg 0V). Then look at the symetry of the circuit. There is no reason to assume that the negative terminal of the battery is also 0V with respect to earth.

    You are on the right lines.

    What would happen if the earth connection was made at the positive terminal of the battery instead of point B? eg What would be the voltage at A and B if C was 0V?
    Last edited: Jan 19, 2013
  4. Jan 19, 2013 #3


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    Actually you don't. The ends of the 10K are connected to the battery terminals so you know the voltage across the 10K resistor is 6V.
  5. Jan 19, 2013 #4
    Right, I see that now. So current across the 10k Ohm resister is 0.6mA.

    I guess my biggest problem right now is not understanding how the ground affects the flow of current:

    If the ground were not there, I would say that the current flows from right to left across all resistors, and back up into the battery. After putting the ground in, this remains true for the single 10k Ohm resister, but I am unsure about the others.

    Clearly the current flows into the 1k Ohm resistor on right towards the ground. From there, though, I guess I'm just not sure. If I read my book right, no current flows into or out of a ground, so would the current continue to behave as if the ground were not present?

    Does the ground not affect the flow? Does it only stipulate that the voltage goes to zero between those 2 1k Ohm resistors?

    Also, with a ground present, doesn't that also imply that the battery voltage potential is measured in terms of ground? (i.e. +6v as compared to ground of 0v)
  6. Jan 19, 2013 #5


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    Good news...it doesn't.

    There is only one connection between the circuit and ground so no current can flow from point B to ground. Therefore the current through both 1k resistors is the same.

    You can choose to measure voltages from where ever you like as long as you are consistent and remember to specify it in the answer. In this case it would be normal to choose Earth or Point B as the reference 0V but you don't have to. Have a go using point B and let us know what you get for A and C

    Suppose I used the +ve terminal of the battery as my reference point. Then...

    C = 0V (w.r.t battery +ve )
    B = Earth = -3V (w.r.t battery +ve)
    A = -6V (w.r.t battery +ve)
  7. Jan 19, 2013 #6
    After going out for dinner with the wife, a lightbulb popped on!

    I think... and correct me if I'm wrong... that the voltage at C is +3v, and the voltage at A is -3v.

    This would satisfy the total electric potential gain of +6v across the battery, AND 0v at ground (at point B).
  8. Jan 20, 2013 #7


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    Just make sure to specify that the voltages you give as your answer are "with respect to Earth (Point B)"

    Might be interested in this..

    Modern cars have the -ve terminal of the 12V battery connected to the chassis. This is called a "negative earth" system because the chassis metal work is assumed to be connected to earth. Wire and switches are used to send power to the lights and the chassis is used to return the current to the battery.

    Prior to 1970 some cars were made that had a "positive earth system". The +ve terminal of the battery was connected to the chassis. This works just as well. The chassis supplies current to the lamps which returns to the battery -ve via switches and wire.

    A problem occured when you wanted to help someone jump start their car and the systems were different. If you used jump leads to connect chassis to chassis and battery +ve to battery +ve then you accidentally shorted out one of the car batteries. A car battery can deliver several hundred amps so this sometimes ended with melted jump leads or fire.
  9. Jan 20, 2013 #8
    Thank you for your help! I just need to get used to the intuition that battery voltage is simply the potential from one side to the other, and not necessarily related to ground.

    That's quite interesting about car batteries... I remember growing up hearing a few concerned "adults" argue over the proper way to connect jumper cables and I never knew why, since it's always been very simple for me and my newer cars. Now I know :)
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