j1 Ω 102 -j10 HH 102 ww x/0° V ZL Question 3) Since the amplitude value of the above circuit de voltage source is x=5V, what is the value of the ZL load in the circuit for the maximum average power and what is the maximum average power of this load? Question 3) The amplitude of the voltage source is x=5V, then find ZL for the maximum average power transfer and calculate its maximum average power. O a. ZL=0.4+ j0.2 Ohm, Vth=3.00+j1.00 V, Pmax = 3.12 W O b. ZL=0.4+ j0.2 Ohm, Vth=25.00+j35.00 V, Pmax=578.12 W O c. ZL=0.4 + j0.2 Ohm, Vth=30.00+j10.00 V, Pmax = 312.50 W O D. ZL=0.4+ j0.2 Ohm, Vth=55.00+j35.00 V, Pmax=1328.12 W O to. ZL=0.4-j0.2 Ohm, Vth=3.00+j1.00 V, Pmax = 3.12 W O f. ZL=0.4 - j0.2 Ohm, Vth=3.50+j10.00 V, Pmax = 35.08 W O g. ZL=0.4-j0.2 Ohm, Vth=30.00+j10.00 V, Pmax = 312.50 W Oh. ZL=0.4 - j0.2 Ohm, Vth=55.00+j35.00 V, Pmax=1328.12 W 87% 8

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Circuit theory
j1 Ω
102
-j10
44
192
ww
x/0° V
ZL
Question 3) Since the amplitude value of the above circuit de voltage source is x=5V, what is the value of the ZL load in the circuit for
the maximum average power and what is the maximum average power of this load?
Question 3) The amplitude of the voltage source is x=5V, then find ZL for the maximum average power transfer and calculate its
maximum average power.
O a. ZL=0.4+ j0.2 Ohm, Vth=3.00+j1.00 V, Pmax = 3.12 W
O b. ZL=0.4+ j0.2 Ohm, Vth=25.00+j35.00 V, Pmax=578.12 W
O c. ZL=0.4 + j0.2 Ohm, Vth=30.00+j10.00 V, Pmax = 312.50 W
O D. ZL=0.4 + j0.2 Ohm, Vth=55.00+j35.00 V, Pmax=1328.12 W
O to. ZL=0.4-j0.2 Ohm, Vth=3.00+j1.00 V, Pmax = 3.12 W
O f. ZL=0.4-j0.2 Ohm, Vth=3.50+j10.00 V, Pmax = 35.08 W
O g. ZL=0.4-j0.2 Ohm, Vth=30.00+j10.00 V, Pmax = 312.50 W
Oh. ZL=0.4 - j0.2 Ohm, Vth=55.00+j35.00 V, Pmax = 1328.12 W
87%
Transcribed Image Text:j1 Ω 102 -j10 44 192 ww x/0° V ZL Question 3) Since the amplitude value of the above circuit de voltage source is x=5V, what is the value of the ZL load in the circuit for the maximum average power and what is the maximum average power of this load? Question 3) The amplitude of the voltage source is x=5V, then find ZL for the maximum average power transfer and calculate its maximum average power. O a. ZL=0.4+ j0.2 Ohm, Vth=3.00+j1.00 V, Pmax = 3.12 W O b. ZL=0.4+ j0.2 Ohm, Vth=25.00+j35.00 V, Pmax=578.12 W O c. ZL=0.4 + j0.2 Ohm, Vth=30.00+j10.00 V, Pmax = 312.50 W O D. ZL=0.4 + j0.2 Ohm, Vth=55.00+j35.00 V, Pmax=1328.12 W O to. ZL=0.4-j0.2 Ohm, Vth=3.00+j1.00 V, Pmax = 3.12 W O f. ZL=0.4-j0.2 Ohm, Vth=3.50+j10.00 V, Pmax = 35.08 W O g. ZL=0.4-j0.2 Ohm, Vth=30.00+j10.00 V, Pmax = 312.50 W Oh. ZL=0.4 - j0.2 Ohm, Vth=55.00+j35.00 V, Pmax = 1328.12 W 87%
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