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Unformatted text preview: problem. A) 1.64A B) 3.29A C) 0.91A D) 2.13A E) None of the above 6 Problem 9 Given ZL = 100 + 100j, N2 =90, N1 =10, find V. N2 ZL + 5A A) B) C) D) E) N1 V − 7.07V ∠ − 135o 63.64V ∠45o 63.64V ∠ − 135o 7.07V ∠45o None of the above Problem 10 Find X such that the maximum power transfer constraint is satisfied. −2jΩ 6jΩ 5jΩ 2jΩ − 5j(V ) + 6jΩ −4jΩ A) B) C) D) E) 1 − 2.5jΩ 1 + 2.5jΩ 2 − 2.5jΩ 2 + 2.5jΩ None of the above X 1Ω (5 + 5j)(A) 5 Problem 7 Find the Norton equivalent current source between A and B. R R R R R R A R I I R B R R A) -2.28I(A) B) -1.24I(A) C) -3.21I(A) D) -6.42I(A) E) None of the above Problem 8 Find N2 and X such that maximum power is delivered to the 1000Ω resistor. 10Ω + 5V − A) N2 = 50, X = −16 B) N2 = 10, X = −16 C) N2 = 10, X = −0.64 D) N2 = 2, X = −0.64 E) None of the above 0.16jΩ 5 : N2 jX 1000Ω 3 Problem 3 Find R that satisfies the maximum power transfer constraint. 10Ω + 100V − 20Ω 20Ω R + 100V − 3A 20Ω 10Ω A) B) C) D) E) 14Ω 15Ω 16.33Ω 17.46Ω None of the above Problem 4 Find V1 . 2Ω + 5V − 2Ω A + V1 2Ω − 6V1 − + 6V B A) B) C) D) E) -1.22V 1.22V -1.57V 1.57V None of the above 2 Problem 1 Find the Thevenin equivalent voltage between A and B (VAB ). −3jΩ A 20jΩ + 3jΩ V0 − 5∠0 2V0 + − B A) 15j V B) -20j V C) 20j V D) -15j V E) None of the above Problem 2 delivered Find the average power associated with the 6A current source between A and B. 2Ω A assume rms values 10jΩ 2jΩ 3A∠0o 6A∠0o B A) -23.86W B) 23.86W C) 28.71W D) -28.71W E) None of the above 5Ω 5Ω 18. Determine Z so that maximum power is transferred to it and calculate this 5Ω power given that the source voltage is 10 V peak value. j2 Ω + 10∠0° V Z – 1:2 Solution: We will determine TEC as seen 5Ω by Z. On open circuit, the currents are as shown. From KVL: 10∠0° – 5I/2 + 5I/2 = I/2 5Ω VTh. In This particular problem, the voltages across the 5 Ω resistors cancel out. Hence, VTh = 10∠0° V peak value I/2 + I I/2 j2 Ω + – 10∠0° V VTh – When Z is replaced by a short 1:2 circuit, the currents are as shown. From KVL: 10∠0° ...
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