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2240 HOMEWORK #6 probs 1-3 solution N. Cotter F13 EX: Find the Thevenin equivalent circuit at terminals a-b. 1) Find the Thevenin equivalent circuit at terminals a-b. vx must not appear in your solution. The expression must not contain more than circuit parameters α, R1, R2, R3, and is. Note: 0 < α < 1. 2) Find the Norton equivalent of the circuit in problem 1. 3) For the circuit in problem 1, assume the following component values: is = 0.4 mA, R1 = 10 kΩΩ, R2 = 2 kΩΩ, R3 = 36 kΩΩ, α = 2 a) Calculate the value of RL that would absorb maximum power. b) Calculate that value of maximum power RL could absorb. SOL'N: 1) SOL'N: 2) SOL'N: 3.a) The maximum power transfer occurs when the load resistance equals the Thevenin equivalent resistance, which may be shown by writing an expression for the power as follows and setting the derivative with respect to RL equal to zero: p = iv = vTh v R ⋅ Th L RTh + RL RTh + RL dp = 0 ⇒ RL = RTh dRL From the answers to (1) and (2) above, we calculate the RTh for the values given. RTh = ( R1 + R2 ) R3 (10 kΩ + 2 kΩ) 36 kΩ = R3 + (1 − α ) ( R1 + R2 ) 36 kΩ + (1 − 2 ) (10 kΩ + 2 kΩ ) or RTh = (12 kΩ) 36 kΩ = 12 kΩ ⋅ 36 kΩ = 18kΩ 36 kΩ −12 kΩ Thus, RL = RTh = 18kΩ 12 kΩ 2 b) When RTh = RL, the formula for power always simplifies as follows: 2 vTh RTh 2 vTh pmax = = ( RTh + RTh )2 4RTh Using the vTh formula from (1) with values given, we get the following value: vTh = is ( R1 + R2 ) R3 0.4 mA (10 kΩ + 2 kΩ ) 36 kΩ = R3 + (1 − α ) ( R1 + R2 ) 36 kΩ + (1 − 2 ) (10 kΩ + 2 kΩ ) or vTh = is RTh = 0.4 mA(18kΩ) = 7.2 V Thus, 2 vTh (7.2 V)2 pmax = = = 0.72 mW . 4RTh 4(18kΩ)