Maximum Average Power Transfer

The maximum power transfer theorem states that, to obtain maximum external power from a source with a finite internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals.

RL= Re{ZTh} and XL = - Im{ZTh}

The maximum power in this case:

Pmax =  

Where V2Th and I2N represent the square of the sinusoidal peak values.

We’ll next illustrate the theorem with some examples.

Example 1

R1 = 5 kohm, L = 2 H, vS(t) = 100V cos wt, w = 1 krad/s.

a) Find C and R2 so that the average power of the R2-C two-pole will be maximum

b) Find the maximum average power and the reactive power in this case.

c) Find v(t) in this case.

The solution by the theorem using V, mA, mW, kohm, mS, krad/s, ms, H, m F units:v

a.) The network is already in Thévenin form, so we can use the conjugate form and determine the real and imaginary components of Z_Th:

R₂ = R₁ = 5 kohm; wL = 1/w C = 2 ® C = 1/w²L = 0.5 mF = 500 nF. 

b.) The average power: 

P_max = V²/(4*R₁) = 100²/(2*4*5) = 250 mW

 The reactive power: first the current: 

I = V / (R₁ + R₂ + j(wL – 1/wC)) = 100/10 = 10 mA

Q = - I²/2 * X_C = - 50*2 = - 100 mvar

c.) The load voltage in the case of maximum power transfer: 

V_L = I*(R₂ + 1/ (j w C ) = 10*(5-j/(1*0.5)) =50 – j 20 = 53.852 e ^{-j 21.8°} V

and the time function: v(t) = 53.853 cos (wt – 21.8°) V