The amplitude of the voltage source is =54V, then find ZL for the maximum average power transfer and calculate its maximum average power. Which option is true?

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The amplitude of the voltage source is =54V, then find ZL for the maximum average power transfer and calculate its maximum average power. Which option is true?
ZL=0.4+ j0.2 Ohm, Vth=32,40+j10,80
V, Pmax = 364,50 W
ZL=0.4+ j0.2 Ohm, Vth=270,00+j378,00 V, Pmax=67432,50 W
ZL=0.4+ j0.2 Ohm, Vth=324,00+j108,00 V, Pmax=36450,00 W
ZL=0.4+ j0.2 Ohm, Vth=594,00+j378,00 V, Pmax=154912,50 W
ZL=0.4 - j0.2 Ohm, Vth=32,40+j10,80 V, Pmax = 364,50 W
L=0.4 - j0.2 Ohm, Vth=37,80+j108,00 V, Pmax=4091,51 W
ZL=0.4 - j0.2 Ohm, Vth=324,00+j108,00 V, Pmax = 36450,00 W
ZL=0.4-j0.2 Ohm, Vth=594,00+j378,00 V, Pmax=154912,50 W
Transcribed Image Text:ZL=0.4+ j0.2 Ohm, Vth=32,40+j10,80 V, Pmax = 364,50 W ZL=0.4+ j0.2 Ohm, Vth=270,00+j378,00 V, Pmax=67432,50 W ZL=0.4+ j0.2 Ohm, Vth=324,00+j108,00 V, Pmax=36450,00 W ZL=0.4+ j0.2 Ohm, Vth=594,00+j378,00 V, Pmax=154912,50 W ZL=0.4 - j0.2 Ohm, Vth=32,40+j10,80 V, Pmax = 364,50 W L=0.4 - j0.2 Ohm, Vth=37,80+j108,00 V, Pmax=4091,51 W ZL=0.4 - j0.2 Ohm, Vth=324,00+j108,00 V, Pmax = 36450,00 W ZL=0.4-j0.2 Ohm, Vth=594,00+j378,00 V, Pmax=154912,50 W
j1 Ω
1Ω
-j1 Ω
1Ω
+1
+ )x/0° V
Z₁
Transcribed Image Text:j1 Ω 1Ω -j1 Ω 1Ω +1 + )x/0° V Z₁
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