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Calculate voltage drop by using kirchoff's law

  1. Jun 15, 2010 #1
    1. The problem statement, all variables and given/known data

    The circuit has a battery Voltage of 30V, a resistor of R1(8ohms), R2 (2Ohms) and R3 (4ohms) in series. Calculate the voltage drop over R1

    2. Relevant equations

    Please give me the proper equation of this? The one i've got says you take 30-8I-2I-4I=0
    Wich gives you current of 2.142 and not the voltage.

    3. The attempt at a solution

    Now what i tried doing is this.
    30V-8-2-4=0
    30V=14
    V=14/30
    V=0.466
    but i'm sure its wrong.
     
  2. jcsd
  3. Jun 15, 2010 #2

    vk6kro

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    Science Advisor

    You have the correct current, although you could just add the resistor values up using mental arithmetic and work out the current using Ohm's Law.

    This current flows through an 8 ohm resistor, so what is the voltage across the resistor?
     
  4. Jun 16, 2010 #3
    ok so what you are saying is that if i use ohms law V=I*R
    I will have to use the 8 ohm resistor times by the current of 2.142= 17.136V
    Is this the voltage after it went through the resistor or is this the amount of voltage drop?
     
  5. Jun 16, 2010 #4

    vk6kro

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    Science Advisor

    The voltage across a resistor is equal to the current through it times the resistance of that resistor.

    So, the 8 ohm resistor times the current of 2.142 Amps = 17.136 Volts.

    If you do this for all the resistors, you will see that they add up to the supply voltage of 30 volts.

    You should also see that the bigger resistors drop more of the the available voltage than the smaller ones, exactly in proportion to their resistance.
     
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