MAXIMUM POWER TRANSFER IN AC CIRCUITS

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POWER IN AC CIRCUITS

THREE PHASE NETWORKS

We have already seen that an AC circuit can (at one frequency) be replaced by a Thévenin or Norton equivalent circuit. Based on this technique, and with the Maximum Power Transfer Theorem for DC circuits, we can determine the conditions for an AC load to absorb maximum power in an AC circuit. For an AC circuit, both the Thévenin 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 reactance cancels the reactance of the Thévenin impedance. Consequently, for maximum power transfer, the Thévenin and load reactances must be equal in magnitude but opposite in sign; furthermore, the resistive parts -according to the DC maximum power theorem- must be equal. In another words the load impedance must be the conjugate of the equivalent Thévenin impedance. The same rule applies for the load and Norton admittances.

R_L= Re{Z_Th} and X_L = – Im{Z_Th}

The maximum power in this case:

P_max =

Where V²_Th and I²_N 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, v_S(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

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