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## Homework Statement

Determine the value of RL that will draw the maximum power from

the rest of the circuit shown below. Calculate the maximum power. (This problem was included in our exam this week, and I don't really recall the exact values of the parameters, but I'm pretty sure that the circuit looks more or less like this.)

## Homework Equations

V=IR

P=I

^{2}R

KVL

## The Attempt at a Solution

*Computing for V*_{ab}:mesh 1:

-10+5I

_{1}+15(I

_{1}-I

_{3})=0

but I

_{3}=0, so I

_{1}=2A

mesh 2:

I

_{2}=3V

_{x}

but Vx=-6I

_{3}, and I

_{3}=0, so I

_{2}=0

Since no voltage exists in the 3 and 6 ohm resistors, I shorted them and V

_{ab}will now be equal to the voltage across the 15 ohm resistor. (Will it? I'm not really sure about this)

Therefore,

V

_{5ohm}=V

_{ab}=15I

_{1}=15(2)

V

_{ab}=30V

*Computing for R*_{ab}:mesh 1:

20I

_{1}-15

_{3}=0 --> eqn.1

mesh 2:

I

_{2}=3Vx

but V

_{x}=-6I

_{3}

I

_{2}=3(-6I

_{3})

I

_{2}+18I

_{3}=0 --> eqn.2

mesh 3:

-15I

_{1}-3I

_{2}+24I

_{3}=0 --> eqn. 3

I

_{1}=-0.011A

I

_{2}=0.270A

I

_{3}=-0.015A

I=-I

_{3}=-(-0.015)

I=0.015A

R

_{ab}=1/I=1/0.015

**R**

_{ab}=66.667ohms

*Computing for the maximum power:*P

_{max}=I

^{2}R

_{ab}=(30/(66.667+66.667))

^{2}(66.667)

**P**

_{max}=3.375WThat is what I did. Sirs, did I solve it the right way? I'm especially uncertain about how I solved V

_{ab}.

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