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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 ThÃ©venin
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
ThÃ©venin 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 ThÃ©venin impedance.
Consequently, for maximum power transfer, the ThÃ©venin 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 ThÃ©venin 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 V^{2}_{Th } and I^{2}_{N} represent the square of the sinusoidal peak values.

Weâ€™ll next illustrate the theorem with some examples.

R_{1} = 5 kohm, L = 2 H, v_{S}(t) = 100V cos
wt,
w
= 1 krad/s.

a) Find C and R_{2} so that the average power of the R_{2}-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 ThÃ©venin form, so we can use the conjugate form and determine the real and imaginary components of Z_{Th}:

R_{2} = R_{1}
= 5 kohm; wL = 1/w C = 2 Â®
C = 1/w^{2}L = 0.5 mF = 500 nF.

b.) The average power:

P_{max} = V^{2}/(4*R_{1}) = 100^{2}/(2*4*5)
= 250 mW

The reactive power: first the current:

I = V / (R_{1} + R_{2} + j(wL â€“ 1/wC)) = 100/10 = 10 mA

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

V_{L} = I*(R_{2} + 1/ (j w C ) = 10*(5-j/(1*0.5)) =50 â€“ j 20 = 53.852 e ^{-j 21.8}^{Â°} V

and the time function: v(t) = 53.853 cos (wt â€“ 21.8Â°) V

V:=100;

om:=1000;

{a./} R2b:=R1;

C2:=1/sqr(om)/L;

C2=[500n]

{b./} I2:=V/(R1+R2b);

P2m:=sqr(abs(I2))*R2b/2;

Q2m:=-sqr(abs(I2))/om/C2/2;

P2m=[250m]

Q2m=[-100m]

{c./} V2:=V*(R2b+1/j/om/C2)/(R1+R2b);

abs(V2)=[53.8516]

** ****Example
2**

v_{S}(t)
= 1V cos w t, f = 50 Hz,

R_{1} = 100 ohm, R_{2}
= 200 ohm, R = 250 ohm, C = 40 uF, L
= 0.5 H.

a.) Find the power in the load R-L

b.) Find R and L so that the average power of the R-L two-pole will be maximum.

First we have to find the ThÃ©venin generator which we will substitute for the circuit to the left of the nodes of the R-L load.

The steps:

1. Remove the load R-L and substitute an open circuit for it

2. Measure (or compute) the open circuit voltage

3. Replace the voltage source with a short circuit (or replace the current sources by open circuits)

4. Find the equivalent impedance

Use V, mA, kohm, krad/s, mF, H, ms units!

And finally the simplified circuit:

Solution for power: **I**
= **V**_{Th }/(**Z**_{Th}
+ R + **j** w L) = 0.511/ (39.17 + 250 â€“ **j **32.82 + **j** 314*0.5)

We find the maximum power if

hence Râ€™ = 39.17 ohm and Lâ€™=104.4 mH.

The maximum power:

I_{max} = 0.511/(2*39.17) = 6.52 mA and

Vs:=1;

om:=100*pi;

va:=Vs*replus(replus(R2,(1/j/om/C)),(R+j*om*L))/(R1+replus(replus(R2,(1/j/om/C)),(R+j*om*L)));

abs(va)=[479.3901m]

PR:=sqr(abs(va/(R+j*om*L)))*R/2;

QL:=sqr(abs(va/(R+j*om*L)))*om*L/2;

PR=[329.5346u]

QL=[207.0527u]

{b./} Zb:=(replus(replus(R1,R2),1/j/om/C));

abs(Zb)=[51.1034]

VT:=Vs*replus(R2,1/j/om/C)/(R1+replus(R2,1/j/om/C));

VT=[391.7332m-328.1776m*j]

abs(VT)=[511.0337m]

R2b:=Re(Zb);

Lb:=-Im(Zb)/om;

Lb=[104.4622m]

R2b=[39.1733]

Here
we used TINAâ€™s special function *replus*
to find the parallel equivalent of two impedances.

POWER IN AC CIRCUITS | THREE PHASE NETWORKS |