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1250 EX: LECTURE #3-5 correction The following are corrected equations for the example problem in video lecture http://youtu.be/fUqX62tbrxE on finding Thevenin equivalent resistance using an external voltage source. Here is the circuit: SOL'N: We turn off the independent source, 68 V, which becomes a wire, but we keep the dependent source turned on. We then attach a 1 V source to the output and determine the current flowing into the a terminal, as shown in the following diagram: Note that a wire should be drawn through the 68 V source. We now use the node-voltage method to find Va and Vc, and from those voltages we will determine Iex and then RTh. Because of the 1 V source, we have Va = 1 V. So we only need an equation for Vc. We sum the currents measured flowing out of Vc. Note how we write Ix in terms of node voltages. ⎛ V − 1V ⎞ Vc − 4 ⎜ c ⎝ 6Ω ⎟⎠ Vc − 1V Vc − 0V + + = 0A 6Ω 2Ω 6Ω Now we collect the terms multiplying Vc and put constants on the right side of the equation. ⎛ 1 1 4 1 ⎞ −4 V 1V Vc ⎜ + − + = + ⎝ 6 Ω 2 Ω 12 Ω 6 Ω ⎟⎠ 12 Ω 6 Ω To clear the denominators, we multiply by the common denominator. ⎛ 1 ⎛ −4 V 1V ⎞ 1 4 1 ⎞ Vc ⎜ + − + ⋅12 Ω = ⎜ + ⋅12 Ω ⎟ ⎝ 6 Ω 2 Ω 12 Ω 6 Ω ⎠ ⎝ 12 Ω 6 Ω ⎟⎠ or Vc ( 2 + 6 − 4 + 2 ) = −4 V+2V or 6Vc = −2V or 1 Vc = − V 3 Now we use Kirchhoff's current law to find find Iex. We sum the currents out of the Va node. 1V − Vc 1V + − I ex = 0 A 6Ω 4Ω or 1 1V − − V 1V 3 + = I ex 6Ω 4Ω or 4 1 + A = I ex 18 4 or 9⎞ 17 ⎛ 8 I ex = ⎜ + ⎟A= A ⎝ 36 36 ⎠ 36 Now we find RTh by dividing the voltage of the external source by the current Iex. RTh = 1V 36 = Ω ≈ 2.12 Ω 17 17 A 36