Problem 4-66

Determine the Thevenin Equivalent

First I will do a circuit reduction of the voltage source and the three left-most resistors to a voltage source in series with a resistor. (The intermediate steps will be left as an exercise for the reader. If you cannot do this by now, then you are in over your head: BACK UP and learn the preceeding material first!)

Note that we definitely do not want to continue the circuit reduction to include the 50 KW resistor.

This would cause us to lose ib, which we need to define the dependent source.

First find v_oc. I will use mesh analysis.

Dependent source parameter:

ib = i₁ - i₃

Current source:

20 ib = i₂

20 (i₁ - i₃) = i₂

20 i₁ - i₂ - 20 i₃ = 0 (Equation 1)

Mesh 1:

6.818 K i₁ + 50 K (i₁ - i₃) - 45.45 = 0

56.818 K i₁ - 50 K i₃ = 45.45 (Equation 2)

Mesh 3:

10 K (i₃ - i₂) + 40 K i₃ + 50 K (i₃ - i₁) = 0

- 50 K i₁ - 10 K i₂ + 100 K i₃ = 0 (Equation 3)

Solving yields

i₁ = 3 mA

i₂ = 10 mA

i₃ = 2.5 mA

Thus

v_oc = 40 K i₃

v_oc = 100 V

Now find i_sc. I will use mesh analysis once again.

As we have seen in earlier problems, since the short circuit is in parallel with a resistor (40 KW in this case), the resistor can be removed with no effect on the rest of the circuit.

Dependent source parameter:

ib = i₁ - i₃

Current source:

20 ib = i₂

20 (i₁ - i₃) = i₂

20 i₁ - i₂ - 20 i₃ = 0 (Equation 1)

Mesh 1:

6.818 K i₁ + 50 K (i₁ - i₃) - 45.45 = 0

56.818 K i₁ - 50 K i₃ = 45.45 (Equation 2)

Mesh 3:

10 K (i₃ - i₂) + 50 K (i₃ - i₁) = 0

- 50 K i₁ - 10 K i₂ + 60 K i₃ = 0 (Equation 3)

Solving yields

i₁ = 5.2 mA

i₂ = 4 mA

i₃ = 5 mA

Thus

i_sc = i₃ = 5 mA

and

R_Th = v_oc / i_sc = 100 / 5 m = 20 KW

Therefore the Thevenin equivalent circuit is:

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