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# 4-69 - The correct answer is therefore p R = = ∞ It may...

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Chapter 4, Solution 69. We need the Thevenin equivalent across the resistor R. To find R Th , consider the circuit below. Assume that all resistances are in k ohms and all currents are in mA. 10||40 = 8, and 8 + 22 = 30 1 + 3v o = (v 1 /30) + (v 1 /30) = (v 1 /15) 15 + 45v o = v 1 But v o = (8/30)v 1 , hence, 15 + 45x(8v 1 /30) v 1 , which leads to v 1 = 1.3636 R Th = v 1 /1 = –1.3636 k ohms R Th being negative indicates an active circuit and if you now make R equal to 1.3636 k ohms, then the active circuit will actually try to supply infinite power to the resistor.

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Unformatted text preview: The correct answer is therefore: p R = = ∞ It may still be instructive to find V Th . Consider the circuit below. (100 – v o )/10 = (v o /40) + (v o – v 1 )/22 (1) 22 k Ω 10 k Ω v 1 + v o − 40 k Ω 30 k Ω 0.003v o 1mA 100V + − 22 k Ω v 1 + v o − 40 k Ω 30 k Ω 0.003v o 10 k Ω v o + V Th − [(v o – v 1 )/22] + 3v o = (v 1 /30) (2) Solving (1) and (2), v 1 = V Th = -243.6 volts...
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4-69 - The correct answer is therefore p R = = ∞ It may...

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