# Maximum average power (to calculate circuit)

#### FuriaseFTW

##### Newbie How can i proceed to solve this exercise? I know that pMAX= |VTh|^2 / 8 * RTh

I tried to calculate the Thevenin Resistance, but i didn't manage to do it, because I can't put together anything in parallel or in series so I really don't know how to find it here.

For the Thevenin Voltage I have the same problem, I don't know which "tactics" I could use here to find the Voltage. #### tyassin Hi

I would say:
2. Calculate the voltage over the removed load (40 ohm and the inductor).
3. Calculate the thevenin impedance when the current source is out (open circuit). What impedance does the Load see, two series branches in parallel.
4. Connect the thevenin voltage with thevenin impedance in series with the Load and calculate the power.

##### Super Moderator
Staff member Impedance matching appears to be key here. To transfer maximum power, adjust outgoing impedance (load ZL) equal to incoming which comprises the remainder of circuit.

On that basis I want to say ZL should be about the same as the other component values, but that's not the scientific answer, is it?

#### Ratch Fairly simply problem. Load is looking at the the voltage impedance of z = {- 10j || 40 } || { 30 || 20j } = 9.88235-8.47059j . We want the load to be the conjugate impedance of the source voltage for maximum power transfer, so load impedance should be zc = 9.88235+8.47059j . Adding the source and load impedance together cancels out the reactance and gives the highest circuit current for equal load and source resistances. Total circuit resistance is 2 times 9.88235 =19.7647 . Dividing into source voltage give 5j/19.7647 = 0.252976 j amps . Squaring the amp value and multiplying by zc gives -63244-0.542092j watts. The negative sign shows the load is absorbing power.
Ratch

• Easy peasy

#### _Eduardo_

##### Full Member level 5 How can i proceed to solve this exercise? I know that pMAX= |VTh|^2 / 8 * RTh

I tried to calculate the Thevenin Resistance, but i didn't manage to do it, because I can't put together anything in parallel or in series so I really don't know how to find it here.

For the Thevenin Voltage I have the same problem, I don't know which "tactics" I could use here to find the Voltage.
...
You only need the Thevenin impedance, what is equal to the impedance see by Zload
Zth = (30-j10)//(40+j20) = 20 ohms

#### Ratch Your analysis is wrong. The voltage source shorts out the two pairs of the impedance bridge arms giving the source impedance shown in my post. Besides, your arithmetic is wrong. The numerator and denominator containing different complex values can never result in a real result. The correct result is 1/2 -1/2 i .
Ratch

• Easy peasy

#### tyassin In #1 there is shown a current source.

#### FvM

##### Super Moderator
Staff member In #1 there is shown a current source.
I presume, anyone noticed. A current source doesn't affect circuit impedance and calculation of matched load. In the second step, the current magnitude is used to calculate load power.
--- Updated ---

In addition, I found that the impedance calculation by _Eduardo_ is correct.

You get admittance of upper and lower branch as
Y1 = 0.03 + j0.01
Y2 = 0.02 - j0.01

Parallel circuit admittance and impedance is
Y = 0.05 + j0
Z = 20

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#### Akanimo ZL is 20 ohm.

To find Pavg, make ZL in post #1 equal to 20 ohm and find the current (ILoad) flowing through it using any convenient method. I would use nodal analysis if I were to do it. ILoad and ZL would be used to calculate Pavg.

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#### Akanimo How can i proceed to solve this exercise? I know that pMAX= |VTh|^2 / 8 * RTh

I tried to calculate the Thevenin Resistance, but i didn't manage to do it, because I can't put together anything in parallel or in series so I really don't know how to find it here.

For the Thevenin Voltage I have the same problem, I don't know which "tactics" I could use here to find the Voltage.

View attachment 165591
I have done it in one of the ways available to determine the parameters listed in the question. There are various ways to do it. You could study what I have done and ask questions where you need to.
View attachment IMG_20201129_182107.jpgView attachment IMG_20201129_182036.jpg

#### _Eduardo_

##### Full Member level 5 Your analysis is wrong. The voltage source shorts out the two pairs of the impedance bridge arms giving the source impedance shown in my post.
Yes, the voltage sources shorts . But that's a current source, which represents an open circuit.

Besides, your arithmetic is wrong. The numerator and denominator containing different complex values can never result in a real result. The correct result is 1/2 -1/2 i .
Never say never:
1/Zth = 1/(30-j10) + 1/(40+j20) = (30+j10)/1000 + (40-j20)/2000 = 3/100 + j/100 + 2/100 - j/100 = 5/100 = 1/20

#### Ratch Sorry, my eyes aren't what they used to be. I missed the symbol for the current source and the double slash for the parallel impedance. I thought you were writing 30-10j/40+20j. I will try to post a correct solution soon. Ratch

#### Ratch To the Ineffabble All,

OK, I think I have a solution to this problem. First, I will outline the
procedure. Then I will proceed with the solution step by step. Please check my
calculations and let me know if I go astray.

We need to find the Thevenin/Norton impedance of the network looking from the
load. To do that we need to find the open voltage across the bridge and the
short current through the wire connected across the same points. Dividing the
open voltage by the short current will give the Thevenin/Norton impedance of
the network. Then inserting a load consisting of the conjugate impedance will
assure the maximum power transfer.

Let's begin the calculations. The bridge has 4 nodes and 2 states, top, bottom,
left, right, open and short. We start by defining and calculating the node
voltages, branch currents which will lead to the eventual calculation of the
Thevenin/Norton impedance.

iS = 5 j (*source current*)

zTO = (40-10j)|| (30+20j) = 103/5 + (21 j)/5 (*total bridge open Z*)

zTS = (30||-10j) + (40||20j) = 11 + 7 j (* total bridge short Z*)

vTO = zTO *iS = -21 + 103 j (*total open bridge voltage*)

vTS = zTS*iS = -35 + 55 j (*total short bridge voltage*)

vLO = (40/(40 - 10 j)) vTO = -44 + 92 j (*left open bridge voltage*)

vRO = ((20 j)/(20 j + 30)) vTO = -54 + 22 j (*right open bridge voltage*)

vTHO = vLO - vRO = 10 + 70 j (*Thevenin open voltage*)

vLS = vRS = ((40||20j)/((40||20j) +(30||-10j)))vTS = -80 + 40 j (*left/right
short bridge voltage*)

iTHS = ((vTS - vLS)/(-10 j)) - (vLS/40) = 1/2 + (7 j)/2 (*Norton short
current*)

Abs[ iTHS ]= 5/Sqrt (*absolute value of iTHS*)

zTHNOR = (vTHO/iTHS) = 20 (*Thevenin/Norton impedance*)

(Abs[ iTHS]^2)* zTHNOR = 250 (*maximum power*)

So, the Thevenin/Norton impedance is 20 ohms resistance, the optimum load for
max power transfer is 20 ohms resistance and the max power delivered is 250 watts.

Ratch