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Exercise 3.10.1AboutWhat value of the load resistor R(a)will extract the maximum amount of power from the circuit below, and how much power willthat be?6Ωww-4Ω4ΩRLb8 Ω) 3 AFeedback?ww 2. For the circuit shown in Exercise 3.10.1, use successive source transformations to reduceeverything to the left of the load resistor into a circuit consisting of justvoltage sourceand a series resistor. What are the correct values of this voltage source and series resistor?one

Question

These are two parts to the ONE problem. The number 2 is just a follow up question.

Exercise 3.10.1
About
What value of the load resistor R
(a)
will extract the maximum amount of power from the circuit below, and how much power will
that be?
6Ω
ww-
4Ω
4Ω
RL
b
8 Ω
) 3 A
Feedback?
ww
help_outline

Image Transcriptionclose

Exercise 3.10.1 About What value of the load resistor R (a) will extract the maximum amount of power from the circuit below, and how much power will that be? 6Ω ww- 4Ω 4Ω RL b 8 Ω ) 3 A Feedback? ww

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2. For the circuit shown in Exercise 3.10.1, use successive source transformations to reduce
everything to the left of the load resistor into a circuit consisting of just
voltage source
and a series resistor. What are the correct values of this voltage source and series resistor?
one
help_outline

Image Transcriptionclose

2. For the circuit shown in Exercise 3.10.1, use successive source transformations to reduce everything to the left of the load resistor into a circuit consisting of just voltage source and a series resistor. What are the correct values of this voltage source and series resistor? one

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check_circleAnswer
Step 1

Figure 1 shows the Thevenin resistance at the terminals a-b. (By removing load resistor and current source as open circuit).

6Ω a
4Q
4Ω
RTh
b
29
8 Ω;
Figure
From Figure 1, the Thevenin resistance is calculated as follows
RT(2+4+4n)|| 8 2]+6
=[102||82+69
Ω18Ω|+
- 4.44 Ω+6Ω
10.44
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Image Transcriptionclose

6Ω a 4Q 4Ω RTh b 29 8 Ω; Figure From Figure 1, the Thevenin resistance is calculated as follows RT(2+4+4n)|| 8 2]+6 =[102||82+69 Ω18Ω|+ - 4.44 Ω+6Ω 10.44

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Step 2

Figure 2 shows the Thevenin voltage at the terminals a-b.

V1
V2
4Q
4Ω
65
w a
ξ2Ω
8S
3A
Ω.
Th
b
Figure 2
From Figure 2, the node voltage V2 = VTh
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V1 V2 4Q 4Ω 65 w a ξ2Ω 8S 3A Ω. Th b Figure 2 From Figure 2, the node voltage V2 = VTh

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Step 3

From Figure...

Apply KCL at node V.
и. и
-VT
Th
=3
6
4
VD3
1
1
Th
6
4
4
0.4167V 0.25V
=3
Th
(1)
or V
Th
Apply KCL at node V,
Vп — И
0
Th
Th
4
1
1
1
V 0
Th
4
-0.25V0.375V 0
(2)
Th
On solving equation (1) and (2)
V 12 V
V 8 V
Th
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Image Transcriptionclose

Apply KCL at node V. и. и -VT Th =3 6 4 VD3 1 1 Th 6 4 4 0.4167V 0.25V =3 Th (1) or V Th Apply KCL at node V, Vп — И 0 Th Th 4 1 1 1 V 0 Th 4 -0.25V0.375V 0 (2) Th On solving equation (1) and (2) V 12 V V 8 V Th

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