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Solve current when only power and resistance is given

max_torch

Feb 9, 2014
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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?
 

Harald Kapp

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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...
 

max_torch

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

Ratch

Mar 10, 2013
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Here we have a circuit:
View attachment 21085
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?

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
 

max_torch

Feb 9, 2014
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One equation sums up the power
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?
 

Ratch

Mar 10, 2013
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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?

The power supplied by the voltage source equals the power dissipated by the three resistors.

Ratch
 

Ratch

Mar 10, 2013
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I2 = 5.61
I3 = 106.866
RL = 2.5198
correct?

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
 

Laplace

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

Laplace

Apr 4, 2010
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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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