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Electric Circuits

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data
    The question is given in the thumbnail...

    2. Relevant equations
    R = resistance P = power V = voltage I = current

    R = V / I
    P = V * I
    P = I * R
    P = V^2 / R

    3. The attempt at a solution
    The attempt at a solution is also given in the thumbnail...

    I know that the total voltage is 3V and after resistor 1: 3V - 1.8V that there will be 1.2 volts remaining. So i'm assuming from here (since I was given the current for resistor 2) that the 1.2 volts carries over to resistor 3, allowing me to determine the resistance r = 1.2v / 0.06a = 20.

    Here is where i'm stuck. With resistor 4, how do I know whether the volts or the current is the value that carries over? I know it has to be one of those two values seeing as how resistor 4 gives you its resistance, so if I were to add the voltage or current to it, I could calculate everything I need. So the question stands, is it the 1.2 volts that carries over to resistor 4, or the 0.06 amps?
     

    Attached Files:

  2. jcsd
  3. Oct 19, 2015 #2

    SteamKing

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    The total current going into the part of the circuit with parallel resistors must equal the total current coming out. The voltage drop in each parallel branch must be the same.
     
  4. Oct 19, 2015 #3
    Ahh okay. So for the last two columns of the chart:

    Resistor 3: 20Ω, 0.06A, 1.2V
    Resistor 4: 12Ω, 0.06A, 0.6V

    And then the power would be Ptotal = Vtotal * I total
    P = 3V * 0.30A
    P = 0.90 watts
    Therefore the power of the circuit would be 0.90 watts. I believe this is correct, yes?
     
  5. Oct 19, 2015 #4

    SteamKing

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    How did you get R3 = 20Ω ?
     
  6. Oct 19, 2015 #5
    I know that the amperage had to be 0.06A because the circuit had a total of 0.3A and resistor 2 had 0.24A going through it, so 0.3 - 0.24 = 0.06A for R3. Also after R1, the remaining voltage was 1.2V which was carried to both R2 and R3. I used 1.2V / 0.06A to get a Ω of 20.

    Im only assuming that this was correct.
     
  7. Oct 19, 2015 #6

    SteamKing

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    In the branch with R3 and R4, you must use the equivalent resistance Req in the equation V = I*R.

    For this branch, Req = R3 + R4 = R3 + 12Ω
     
  8. Oct 19, 2015 #7
    But if you dont know what R3 is then how are you supposed to find the equivalent??
     
  9. Oct 19, 2015 #8

    SteamKing

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    You know what the current I is and you know that the voltage drop in each branch is the same. You can calculate the voltage drop in the branch which contains R2 and use this voltage drop to calculate R3, as described above.
     
  10. Oct 19, 2015 #9
    Im not quite following....because if I do that i'd be getting the exact same answer:

    The amperage in R2 is 0.24, and its resistance is 5. That means that 0.24 * 5 = 1.2 volts. Which would mean that 1,2 volts also goes into R3. In R3 if the voltage is 1.2 and the amperage is 0.06, you would plug them in. R = V / I R = 1.2volts / 0.06amps resistance = 20.

    Im not sure where you were going with the R3 + R4 business...I didnt learn that at all during my lesson. But how would this NOT be the answer?
     
  11. Oct 19, 2015 #10

    SteamKing

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    You're not following because the R you are calculating is the Req of the branch.

    Remember, Req = 20Ω = R3 + R4, and R4 = 12Ω. So what is R3 ?
     
  12. Oct 19, 2015 #11
    So R3 would be 8 then...?
     
  13. Oct 19, 2015 #12

    SteamKing

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    Correct, but more accurately 8Ω.
     
  14. Oct 19, 2015 #13
    I have the final answer but im afraid im not understanding how I got there. You said that Req was equal to 20....but where did that number even come from?
     
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