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Find the average power absorbed by a specific resistor in circuit

  1. Jan 2, 2014 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/a/img850/788/yk72.jpg [Broken]

    Find the average power absorbed by the 2Ω resistor in the circuit (pictured above)

    2. Relevant equations

    Average power = (1/2)(Imax)(Vmax)(cosø)


    current division, voltage division

    3. The attempt at a solution

    So trying to get total impedance, first I get this (2 || -2j):

    http://imageshack.us/a/img30/5500/wpwn.jpg [Broken]

    then this (1-1j + 2j):

    http://imageshack.us/a/img856/1788/u357.jpg [Broken]

    then (1 +1j) || 4

    = (1/1+1j + 1/4)-1

    = (5 + 1j)/(4 + 4j)-1

    = (4 + 4j)/(5 + 1j)

    multiplying by conjugate (5 - 1j)/(5 - 1j) got me

    = (24 + 16j)/24

    Z total = 0.923 + 0.6153j

    Z total = (1.109 ∠ 33.68°)Ω

    I total = (6∠0)/(1.109 ∠ 33.68°)

    I total = (5.41 ∠ -33.68A)

    doing current division, current in the middle branch is:

    I = (5.41 ∠ -33.68)*(4 / [5 + 1j] )

    I = (5.41 ∠ -33.68)*(4∠0 / 5.099 ∠ 11.3)

    I = (4.243 ∠ -45°)A in mid branch

    Then doing current division again where 2 || -2j, the current through the 2Ω would be

    I = (4.243 ∠ -45°)A * (-2j/[2 - 2j] )

    I = (4.243 ∠ -45°)A * (2 ∠ -90 / 2.82 ∠ -45)

    I = (3.009 ∠ -90°)A through the 2Ω

    For the voltage across the 2Ω it's

    V = (6 ∠ 0)*( [1-1j]/[1 + 1j] )

    V = (6 ∠ 0)*(1.41∠-45 / 1.41∠45)

    V = (6∠-90°)V across the 2Ω resistor

    So now I have

    Pavg = (1/2)*(6)(3.009)*cosø

    So my only question is, how to find ø? Is it the angle of total impedance of the whole circuit or just of the 2Ω resistor?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 2, 2014 #2
    I assume that 6|_0 is the value for voltage source. here 6 is the rms value of voltage
    in the branch near this voltage source, impedance is 1+1j as calculated by you. let z1= 1+1j
    The voltage source is connected across this z1.
    let v= 6|_0.
    complex power consumed by the brach is v^2/z1. do it with scientific calculator. since only one(2 ohm) resistor is present in the branch, the real part of the complex power gives the average value of power consumed by the resistor.
  4. Jan 2, 2014 #3


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    Staff: Mentor

    The symbol for the source in your diagram is not very common. Can you confirm that it is a voltage source rather than a current source? Usually an arrow (or arrow-like symbol) included inside a circle or diamond shape implies a current source.

    If it is indeed a voltage source, what's the corresponding symbol for a current source in your symbol system?
  5. Jan 2, 2014 #4
    Yes it's a voltage source. I've edited the starting problem image to show exactly how it looked on the original question now. Is it still RMS value?
  6. Jan 2, 2014 #5


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    Staff: Mentor

    I believe that that would make it a current source. The circle with an arrow inside of it is a standard one for a current source.
  7. Jan 2, 2014 #6
    It still says 'V' as if it were a voltage source though. I'm not too sure what it means, but that's exactly how it's given (along with the arrow though).
  8. Jan 2, 2014 #7


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    Staff: Mentor

    Very strange. It's marked "6 ∠0° V" beside it?
  9. Jan 2, 2014 #8
    Yes. I'll try it over as a current source then and get back to you.
  10. Jan 2, 2014 #9
    Sorry for getting back to this late..

    Okay it looks like it was a typo on the Professor's part or whoever printed it (it's printed on a paper not from a textbook but probably copied over wrong or something. Whatever.)

    Even if it's a current source, the impedance in the circuit is still the same. So,

    Z total is still = (1.109 ∠ 33.68°)Ω assuming I still found this correctly

    total voltage is V = IZ, so

    Vtot = (6 ∠ 0)*(1.109 ∠ 33.68)

    Vtot = (6.654 ∠ 33.68)V

    Then, doing the same methods as before (current and voltage divisions) I got

    I = (4.706∠-11.3)A in middle wire

    I = (3.337∠-56.3)A for the 2Ω

    V = (6.654∠-56.32)V across the 2Ω

    Pavg = (1/2)(6.654)(3.337)cosø

    average power is still be without an angle for cos though, what would it be? The total impedance of the circuit?
  11. Jan 3, 2014 #10


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    Staff: Mentor

    As soon as you have found the current in the 2Ω resistor of interest you are finished. Just I2 R and you're done. :smile: Power doesn't have an angle.

    Whatever means you use to arrive at the voltage across the resistor (or any resistor, for that matter), it must always have the same angle as that resistor's current. So once you have determined one, you have as good as determined the other, just apply Ohm's Law (merely a scaling factor).
    Last edited: Jan 3, 2014
  12. Jan 3, 2014 #11
    Thought there was a difference between power and average power?

    Got 22W for that then
  13. Jan 3, 2014 #12


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    Staff: Mentor

    That is indeed true. Instantaneous power in the resistor does vary sinusoidally, and it does so at double the frequency of the voltage and current sinusoids.

    p(t) = v(t).i(t)
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