1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Power dissipated and current of a battery in a particular circuit

  1. Feb 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Four resistors are connected in a circuit with a battery of unknown emf, as shown below. The battery is ideal, so it has no internal resistance. The current in the 30Ω resistor is 100. milliamps.

    18-p-052.gif

    C) Calculate power dissipated by the 20Ω resistor in watts.
    D)Calculate the current in the battery in milliamps

    2. Relevant equations
    P=V2/R
    V=RI

    3. The attempt at a solution

    Parts A and B of this question involved finding the current in the 60Ω resistor, and then the current in the 80Ω resistor, which I was able to do, getting 50mA and 150mA respectively. I'm not sure what that has to do with C and D. The problem with finding power is I do not know voltage. I thought I could get that easily, knowing I=150mA for the 80Ω resistor (V=IR=4.8V), but this doesn't work out, so I'm guessing I did something wrong there.

    I'm really at a loss, so if somebody could help me step by step through this, it would be a big help.
     
  2. jcsd
  3. Feb 16, 2014 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Uncle Ohm can help you find the emf of the battery. Then you use him again to get the current through the 20 ##\omega## resistor.
    You tried V=IR to get 4.8 Volt where exactly ?
     
  4. Feb 16, 2014 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Welcome to PF!

    Knowing the currents through the 60Ω, 30Ω, 80Ω resistors you are able to calculate the voltages across the resistors, and from them, the voltage across the battery (which is the same as the voltage across the 20Ω resistor).

    ehild
     
  5. Feb 16, 2014 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Ah yes, welcome ! And ##\omega## should read ##\Omega##. Ehild and I crossed quick replies, something that happens quite often I discovered.
     
  6. Feb 17, 2014 #5
    Alright, so I find the voltages of the resistors in parallel, and they are the same. Apparently, I just add the voltage of one of the resistors in parallel (3V) to the voltage of the 80Ω resistor (12V) that they are in series with. Then I have the voltage of the battery. Why is this so? Why do I not have to, say, add 3V+3V+12V, taking into account voltage of every single resistor, even if one pair is in parallel?

    And thank you both for the help, and the welcome. I've just been stuck on this for quite awhile and couldn't find help and an explanation anywhere else.
     
    Last edited: Feb 17, 2014
  7. Feb 17, 2014 #6

    FOIWATER

    User Avatar
    Gold Member

    You can find the equivalent resistance of the entire right hand branch. Keeping in mind you know the current through the 80 ohm resistor, you also know the current through the entire right hand branch. Knowing the current through the entire right hand branch as well as its equivalent resistance helps you find the voltage across that part of the circuit.. which is the voltage across....
     
  8. Feb 17, 2014 #7

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The voltage between two points - as A and B - is equal to the potential difference U(A)-U(B)=3V here. Also the potential difference between B and C is U(B)-U(C)=12V. The potential difference U(A)-U(C)=[ U(A)-U(B)]+[U(B)-U(C)] = 3+12 =15 V. That is the potential difference across the battery.

    ehild
     

    Attached Files:

  9. Feb 17, 2014 #8

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oh oh, almost hairless:
    Form an image of a pond with two outlets weirs to an 0.3 m lower pond. One of the weirs is approximately twice as wide as the other (the 30 ##\Omega## resistor is easier to pass through than the 60 ##\Omega## ). Through the first weir there goes twice as much water as goes through the narrower one. But they both "see" the same single level difference between the two ponds!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Power dissipated and current of a battery in a particular circuit
Loading...