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You have to take into account the voltage across the current source, call it Vi. So the mesh equations are Vi-(2j*2j)+2*I1-24=0,2*I2+(-2j*I2)+(2j*2j)-Vi=0,I2-I1=2j. Solving for I2 gives 22/5+j*16/5. Multiplying by the 2 ohm output resistor gives 44/5+j*32/5 = 10.88/_36°.
Opening up the circuit at the left side of the capacitor gives a Voc of (24+(2*2j) and a Zo of 2 ohms. Using the voltage divider formula we get (24+(2*2j)*{2/(2+(-2j)+2)} = 10.88/_36°.
Actually, using the 'supermesh' technique his equations look correct, and indeed will produce the correct result except he didnt do the routine math correctly.
Solving the first equation for i1 knowing what i2 is (second equation) leads to a direct solution of i1, and that takes us to i2 which then multiplied by the output resistor will give the right answer.
PG:
Just go over your math if you want to use the supermesh technique.
It is probably best to convert everything to complex form and then proceed. That means the current source is 0+2*j for example.
@Ratch: Thank you. It was good to learn an alternative approach to solve that circuit. And while using the Thevenin, I was applying it wrongly! I have solved the problem now. Much obliged.
@MrAl: Thank you. It was in fact an math error which was producing wrong answer. After correcting it, I was able to solve using supermesh technique. Thanks.
Yes, I can't follow what you are doing, but it seems like a very convoluted way of finding the correct output Z. Instead of doing it the Thevenin way, I am going to show you a general immittance theorem that will make life easier. We first find the transfer function. It does not matter which one as long as the load is included in the function. First find I2 using mesh analysis.
So, I2 = (12+24j)/(2+2j+ZL) . Dividing by the 24 volt source we get the transfer function I2/24 = (1/2+j)/(2+2j+ZL) . Now look at the demoninator carefully. It contains both the input impedance of the the network looking into it from the 24 volt source, and also the output impedance looking into the network from the ZL impedance. We are only interested in the output impedance, so set the denominator to zero and find -ZL . -ZL = 2+2j, so 2+2j is the output impedance or Thevenin impedance of the circuit. As you correctly stated, the max power occurs when ZL is the conjugate of the output impedance or ZL = 2-2j. Substituting ZL into the I2 given at the beginning of this paragraph, we get I2 = 3+6j . Squaring the current I2 and multiplying by ZL we get 18+126j . So ZL is dissipating 18 watts of power at its optimum value.
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I have solved the problem now. Much obliged.
