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## Homework Statement

For the following circuit vs = 100 cos(1000 t + 20o) V.

Case 1:

If the load impedance for the circuit is adjusted until maximum average power is

delivered to the load:

a) Find the impedance that should be connected in this case.

b) Find the maximum average power delivered to the load in this case.

Case 2:

If the load is a pure resistive:

a) Find the value of the load resistor that will maximize the average power in the

load.

b) Find the maximum average power delivered to the load in this case.

Case 3:

If the load is replaced with a series combination of 5 Ohms resistor and a 5 μF capacitor

and the one Ohm resistor is replaced with a box called Load2.

a) Find the impedance that should be connected to the location of load2 to

maximize the average power in Load2.

b) Find the maximum average power delivered to the load in this case.

c) If Load2 is a pure resistive element, what should be the value of this load to

maximize the average power in Load2 and what is the value of the maximum

average power in this situation.

## Homework Equations

## The Attempt at a Solution

Case I:

*The value of the 10 mH conductor = ωjl = 1000*10m = j10Ω

*The value of the 3 mH conductor = ωjl = 1000*3m = j3Ω

A)we have to find Thivenin impedance :

- Applying KVL in the first loop to find (ix):

- 100∟20+(25+j10) ix + 5 ix =0

(30+j10) ix =- 100∟20 → i_x=3.162+j0.1=3.162∟1.8 A

→ The dependent voltage source = 15.8∟1.8 V

- Using source transformation:

Current source 1 = 3.162∟1.6 A & parallel to (25+j10) Ω impedance

& Current source 2 = 15.8∟1.8 A & parallel to 1 Ω resistor

- combining parallel impedances to find equivalent impedance Z_th:

(25+j10) ║ 1 ║ j3

→ 1/Z=1/((25+j10))+1/10+1/( j3) →Z_th=0.6∟13= 0.585+j0.135 Ω

→ Z_l= Z_th^* = 0.6∟-13= 0.585-j0.135 Ω

Which will gives maximum power.

B) finding the complete Thevenin circuit to find the value of the power:

- combining parallel current sources:

3.162∟1.6 + 15.8∟1.8 = 19∟4.39 = 19+j1.46 A

- source transformation again to have Thevnin equivalent circuit:

→ V_th= 11.4∟17.4 V

→ P_max=〖(I/√2)〗^2*R_th = 〖(19/√2)〗^2*0.585 = 105.6 W

Case II:

A) The value of the load resistor = √((R_th^2 )+(X_th^2 ))= √(〖(0.585)〗^2+〖(0.135)〗^2 )= 0.6 Ω

B) The maximum power = (V_m^2)⁄(8R_th )= 〖(11.4)〗^2⁄((8)(0.6))=27 W

Case III :

*The value of the 5µF capacitor = (-j )/ωc= - j200Ω

A) we have to find the Thevinin equivalent circuit:

- using parallel/series combinations:

(5-j200) ║ j3 = (j3(5-j200))/(j3+(5-j200)) = 3.04∟89.9 = 0.01+j3.04 Ω

- using source transformation:

Current source 1 = 3.162∟1.6 A & parallel to (25+j10) Ω loud.

& Current source 2 = 5.197∟-88.2A & parallel to (0.01+j3.04) Ω loud.

- combining parallel impedances to find equivalent impedance Z_th

(25+j10) ║ (0.01+j3.04)

→Z_th=((25+j10)(0.01+j3.04) )/((25+j10)+(0.01+j3.04)) →Z_th= 2.9∟84.1 Ω

→ Z_l=2.9∟-84 which will give the maximum power.

B)Finding the complete Thivenin circuit to find the value of the power:

- combining parallel current sources:

3.162∟1.6 + 5.197∟-88.2 = 6.1∟-56.9 A

- Source transformation again to have Thevnin equivalent circuit:

→ V_th= 17.69∟27.2 V

→ P_max= (V_m^2)⁄(8R_th )= 〖(17.69)〗^2⁄((8)(0.298))= 131.26W

C) The value of the load resistor = √((R_th^2 )+(X_th^2 ))= √(〖(0.298)〗^2+〖(2.885)〗^2 )= 2.9 Ω