 #1
 17
 0
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}=3.375W
That is what I did. Sirs, did I solve it the right way? I'm especially uncertain about how I solved V_{ab}.
Attachments

16.9 KB Views: 373
Last edited: