MEL TECH
CIRCTRONICS
MAXIMUM AVERAGE POWER TRANSFER
The maximum power transfer theorem for DC circuit, we can determine the condition for an AC load to absorb maximum power in an AC circuit. For an AC circuit, both the thevenin impedance and the load can have a reactive component. Although these reactances do not absorb any average power, they will limit the circuit current unless the load reactances cancels the reactance of the thevenin impedance. For the maximum power transfer, the thevenin and load reactances must be equal in magnitude but opposite in sign.
If the load is purely real, then RL = √(Rth)^2 + (Xth)^2 = |Zth|
Zth = Rth + jXth ; ZL = RL + jXL
Prove that:
XL = -Xth & RL = Rth
SOLUTION: ( THIS IS BASED ON MY OWN SOLUTION)
P = 1/2 (I)^2 (RL)
I = VTH/Zth + ZL
P = 1/2 |Vth/(Zth+ZL)|^2 (RL)
P = 1/2 |Vth/(Rth+jXth)+(RL+jXL)|^2 (RL)
P = 1/2 |Vth^2 (Rth + jXth + RL + jXL)^-2| (RL)
dP = 1/2 |Vth^2 (Rth + jXth + RL + jXL)^-2| d(RL) + 1/2 (RL) |Vth^2 (dRL) (-2) (Rth + jXth + RL + jXL)^-3|
dP/dRL = 1/2 |Vth^2 (Rth + jXth + RL + jXL)^-2| + 1/2 (RL) |Vth^2 (-2) (Rth + jXth + RL + jXL)^-3| = 0
(1/2 Vth^2) - RL (Vth^2) = 0
(Rth + jXth + RL + JXL)^2 (Rth + jXth + RL + jXL)^3
1 - RL = 0
2 Rth + jXth + RL +jXL
1 = RL
2 Rth + jXth + RL +jXL
Rth + jXth + RL + jXL = 2RL
RL = Rth + j (Xth + XL)
RL - Rth = j (Xth +XL)
(RL - Rth) - j (Xth +XL) = 0
-j (Xth +XL) = 0
Xth = -XL
XL = -Xth
RL - Rth = 0
RL = Rth
Example:
1. Given:
V = 200∠8 V
I = 10∠30 A
Z = 20∠-22 = 18.54 - J7.49ohms
2. Get the Maximum Power in ZL.
P = 1/2 (I)^2 (RL)
P = 1/2 (10)^2 (18.54)
P = 927.2W
Using Thevenin's Theorem, open the current source to get the Zth.
Zth = 8+j6Ω || 5Ω
= (8+j6Ω) (5Ω)
(8+j6Ω)+(5Ω)
Zth = 3.415 + j0.73Ω
In order to get the ZL, get the complex conjugate of Zth...
ZL = 3.415 - j0.73Ω
Get Vth, Using Current Division..
I = (2A) (8-j4Ω) = 0.780 - j0.98
(8-j4Ω)+(5+j10Ω)
Vth = (0.780 - j0.98A) (5Ω)
Vth = 6.247∠-51.34 V
P = 1/2 (I^2) (RL)
P = 1/2 |Vth/(Rth+jXth)+(RL+jXL)|^2 (RL)
P = 1/2 (Vth^2 (RL)) / (2Rth)^2
P = 1/2 (Vth^2 RL) / (4Rth^2) ; RL = Rth
P = Vth^2
8Rth
P = (6.247)^2
8(3.415)
P = 1.43 W