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Problem with electric diagrams.

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known data

    so i've had no real problems with this unit at all until i got to this first question. I had no idea where to start, but i tried working some things out, and couldn't get the right answer. the second question seems to be along the same lines with my problem. when the given amperage is in a parallel circuit I don't know what to do.

    1. What is the EMF of the battery if the current in A is 1.2 A, and the internal resistance of the battery is 0.0833 Ω?

    2. What is the voltage V of the power supply?

    (both diagrams are in the attachment)

    2. Relevant equations

    Vterminal = E +- Ir
    V = IR
    P = IV


    3. The attempt at a solution

    http://dl.getdropbox.com/u/565477/physics.jpg [Broken]

    1. I tried to find the total amperage by doing: 1.2A/6.0Ω = amperage/2Ω. which i got .4, then added it with the other amperage to get 1.6A. I doubt this is proper it was just the only thing i could think to do to find the total amperage. (I've now noticed that they are inversely proportionate, but I still don't know if this is the right approach...)

    2. I then tried to use that amperage I found to find the Voltage, and then plug it into 'Vterminal = E +- Ir', but (I guess obviously) I had no luck.

    Can someone please explain to me how on earth to do that first question, I think I'm just missing some basic piece of knowledge and the second question will be really easy once I know how to do the first one.

    I know the sum of the parallel amperages will give me the total amperage, but I'm just totally stuck with what to do here.

    Thanks...
     

    Attached Files:

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 6, 2009 #2

    vk6kro

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    There does't seem to be any attachment. Could you try again?

    If a battery is in a series circuit, the current flowing is the EMF of the battery divided by the total resistance in the series circuit, including the battery's internal resistance.

    You get the total resistance by just adding up all the series resistances.
    This is just I = V / R which is a variation on one of your formulae. Ohm's Law.

    If you know the voltage across a number of parallel resistors, you can work out the current in each of them by using the same formula above.
     
  4. Sep 7, 2009 #3
    ahhh shoot. i'm sure i had put it in there. i did it two ways this time. thanks for your help..
     
  5. Sep 7, 2009 #4
    ps. i could still use help.
     
  6. Sep 7, 2009 #5

    vk6kro

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    You know the resistance of one resistor and the current through it.
    So, you can work out the voltage across it.

    There is a resistor in parallel with it, so you know the voltage across this resistor.
    So you can work out the current through this resistor.

    Now work out the total current flowing in the bottom resistor (by adding the last two currents).

    So, get the voltage across this resistor.

    Do the same with the internal R of the battery.

    Now, all these voltages added up must equal the EMF of the battery.

    The second problem is similar. Start with what you know and see whatever else you can work out from that.
     
  7. Sep 7, 2009 #6
    i will give it another shot, i think my problem is that i don't really know how the battery formula works i don't think.
     
  8. Sep 7, 2009 #7
    okay so i found the V:
    1.2*6 = 7.2V.
    i then found the parallel amp:
    7.2=2*I=3.6A

    the total amperage is 10.8 amps.

    the total resistance is 1/2+1/6=1/r = 1.5. +.5 + .0833 = 2.0833Ω

    i don't know what to do next. i'm sure i have all the info that I can get from this...i feel like i'm supposed to use Vterminal = E +/- Ir. but i don't actually know or not?
     
  9. Sep 7, 2009 #8
    ps. i got the second question right, thanks to your help. it was pretty easy. it's just this last bit on the other one i'm stuck on.
     
  10. Sep 7, 2009 #9

    vk6kro

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    Who said add up the resistances? You are adding up voltages

    Now work out the total current flowing in the bottom resistor (by adding the last two currents).

    So, get the voltage across this resistor.

    Do the same with the internal R of the battery.

    Now, all these voltages added up must equal the EMF of the battery.
     
  11. Sep 7, 2009 #10
    i did that. the voltage of the parallel part is 7.2V, the bottom one is 5.4V, the resistor is .89V...

    but the right answer to this question is 10V. adding them together doesn't get 10V.. i really need someone to just show me how to do this last step.
     
  12. Sep 7, 2009 #11

    vk6kro

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    The total current is 4.8 amps. (1.2 + 3.6).

    The voltage across the parallel part is 7.2 volts (1.2 * 6 or 3.6 * 2).

    The voltage across the 0.5 ohm resistor is 2.4 volts (0.5 * 4.8 amps)

    The voltage across the 0.0833 resistor is 0.4 volts ( 0.08333 * 4.8)

    Add these up and you get 10 volts.

    I couldn't work out how you got your figures. This problem did not require the formula for terminal voltage (E - IR). It was just Ohms Law. I = V / R.
    Try working through the above example and see if you can understand how it works. You can't plug stuff into formulas and hope you get the right answer. Get back to me if you don't understand.
     
  13. Sep 7, 2009 #12
    i think i just tried it so many times i messed myself up. it's pretty easy, thanks for your help. is there anyway you can check the other question i posted real quick? you're a life saver.
     
  14. Sep 7, 2009 #13

    vk6kro

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    Total current 4 + 2 = 6A
    Voltages 48 v and 24 volts
    Total voltage 72 volts.

    Is that what you got?
     
  15. Sep 7, 2009 #14
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