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(ELEC101)T04Example.pdf
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Example 1: (From 2.33, Introduction to Electrical Engineering by Irwin)
Computer Vo in the network in Fig. 1









Answers to Example 1:


















Fig. 1





Example 2: (From 2.44, Introduction to Electrical Engineering by Irwin)
Use Thevenins theorem to find Vo in the network shown in Fig. 2









Answers to Example 2:
We want to find the Vo for the 3 resistor, so we treat the 3 resistor as the load.

Lets use another method to find the Thevenins equivalent. We short the voltage sources to find the RTH

After short-circuiting all the voltage source and treat 3 as load, we get

Convert the Thevenins equivalent of the 2 sides of Fig 3 to Norton equivalent
























Fig. 2

RTH = 2 // 4 =


From 24V and 2

From 12V and 4




Example 3: (From 2.48, Introduction to Electrical Engineering by Irwin)
Find Io in Fig 3.

Answers to Example 3:
I will use superposition approach to solve this problem.

1.
Consider the 24V source (i.e. Open the 4A source and Short the 36V source)

The circuit reduced to











2.
Consider the 4A source (i.e. Short the 24V source and Short the 36V source)

The circuit reduced to










Note: I4A means the current contributed by the 4A source

3.
Consider the 36V source (i.e. Open the 4A source and Short the 24V source)

The circuit reduced to














You are encouraged to use other methods, e.g. Source Transform, nodal analysis, mesh analysis to solve the problem. You should achieve the same answer.

Fig. 3





Example 4: (From 4.20, Electric Circuits by Nilsson)
Find the Thevenins equivalent circuit with respect to the terminal a, b for Fig 4.








Answers to Example 4:
First, short circuit the 72V source to find the RTH

Find Vab



Subsistute (3) and (1) into (2)



I1 = 4I2 = 2.4A

Vab = 72 . 12I2 = 72 . 7.2 = 64.8V

Thevenins equivalent



Fig. 4

RTH

1.
5 and 20 are in parallel, gives 4


2.
4 and 8 are in series, gives 12


3.
12 and 12 are in parallel, gives 6



RTH = 6

Apply KCL on node Z
I1 + I2 = I3 ...(1)

Apply KVL on Loop 1
5I1 + 20I3 = 72 ...(2)

Apply KVL on Loop 2
5I1 = 12I2 + 8I2 = 20I2
I1 = 4I2 ...(3)