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Solve current when only power and resistance is given

Discussion in 'Electronics Homework Help' started by max_torch, Jul 21, 2015.

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  1. max_torch

    max_torch

    108
    1
    Feb 9, 2014
    Here we have a circuit:
    ckt help.png
    What is the best technique for solving the voltages and currents in the circuit if you are only given power drawn by the load and power generated by the source? I know KVL, KCL, Mesh, Nodal. I tried to do mesh, replacing the voltages with P/I because the voltages are unknown, and I did substitution from the 2nd mesh equation to the first, but then I got a complicated 4th degree equation. I am not sure if what I am doing is correct.
    What is the best approach?
     
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    10,592
    2,362
    Nov 17, 2011
    Try using P=V²/R or P=I²*R.
    Since you know that the voltage across R2 and R3 is the same and you also know that the currrent through R1 equals the sum of currents through R2 and R3 you can set up the necessary equations.

    For example:
    P(R2) = 48W = V(R2^)²/R2 -> V(R2)²=48W*R2
    P(R3)=28.8kW=V(R3)²/R3=V(R2)²/R3 -> 28.8kW=48W*R2/R3 or R2/R3=28.8kW/48W


    Simplify the schematic by replacing R2 and R3 by a single equivalent resistor...
     
    Arouse1973 likes this.
  3. max_torch

    max_torch

    108
    1
    Feb 9, 2014
    0.04 and 48 is not power it is ohms.. sorry if the diagram was not clear
     
  4. Ratch

    Ratch

    1,088
    331
    Mar 10, 2013
    There are three unknowns which can be solved with three equations. There are four solution sets. Only two sets are realizable. One equation sums up the power. The second equates the voltage across R2 and R3. The third equates the R3 resistance and I3 current to the power dissipated in R3. Three equations and four sets of three unknowns, I2, I3, R3 each. Now, write the equations and solve for the unknowns.

    Ratch
     
  5. max_torch

    max_torch

    108
    1
    Feb 9, 2014
    I just want to clarify this section.
    Is this the relationship which states that power generated equals power dissipated?
    Psource = 'P of R1' plus 'P of R2' plus 'P of R3'
    correct?
     
  6. Ratch

    Ratch

    1,088
    331
    Mar 10, 2013
    The power supplied by the voltage source equals the power dissipated by the three resistors.

    Ratch
     
  7. max_torch

    max_torch

    108
    1
    Feb 9, 2014
    I2 = 5.61
    I3 = 106.866
    RL = 2.5198
    correct?
     
  8. Ratch

    Ratch

    1,088
    331
    Mar 10, 2013
    You can check it yourself. Does ((5.61+106.866) ^2) * 0.04 + (5.61^2 ) * 48 + (106.866^2) * 2.5198 = 30614.4 ? What is the second solution? Is RL the same as R3?

    Ratch
     
  9. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    I put together the following spreadsheet to check the correctness of the two solutions. Recommend you do the same.
    EP66H.png
     
  10. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    FYI, these are the two power equations I fed into Maple symbolic algebra engine to solve for the two node voltages:

    EP66E.png
     
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