- #71
Studiot
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- 9
Since I have done all the work for you in the votlage analysis, you could also perform a similar analysis with current instead of voltage.
Allow a current I to flow from A to B through your network and calculate the currents in each resistor in terms of I. (no voltage assumptions are required in this case)
Assemble a total power equation as before using power = (curent in each resistor)2 time the resistance.
Solve for I using the term with the largest coefficient.
Back substitute I to get the total power.
This was my original suggestion about current. I have emboldened the important point.
Just as when working in terms of voltage, current did not appear anywhere
When working in terms of current voltage will not appear at all.
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Here is a start
Let a current I enter the network at A and leave at B.
It is usual to work in terms of conductance (=reciprocal of resistance) for current.
So at the first branch the branch conductances are 0.02 through the 50 ohm and 0.025
Through the other leg.
The current I divides in proportion to the ratio of the conductance of each leg to the total.
So the current in the 50 ohm resistor is
[tex]I\left( {\frac{{0.02}}{{0.02 + 0.025}}} \right)[/tex]
And the current in the other leg is
[tex]I\left( {\frac{{0.025}}{{0.02 + 0.025}}} \right)[/tex]
further subdivision of this leg will yield the other currents
To assemble the power equation we use P = I2R (as we used P = V2/R before).
Can you take it from here?