# Solve current when only power and resistance is given

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

1. ### max_torch

108
1
Feb 9, 2014
Here we have a circuit: 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 KappModeratorModerator

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

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

4. ### 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

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

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

Ratch

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

8. ### 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

1,252
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Apr 4, 2010
10. ### Laplace

1,252
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Apr 4, 2010  