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Problem 1: [12 points] (CIRCLE THE CORRECT ANSWER. NO CREDIT FOR MULTIPLE ANSWERS.) 1) Determine the value of Zeq between a and b if ω is 104 rad/s. 0.1 uF k a) 2200 + j600 Ω b) 2200 – j600 Ω a c) -2200 – j600 Ω k 50 mH d) -2200 + j600 Ω b 2) Find is(t) for i1(t) = 2cos(1000t + 45) A. a) 1.414cos(1000t + 45) A b) 1.414cos(1000t – 45) A c) 1.414cos(1000t ) A d) None of the above 3) The RMS (effective) value for the periodic current shown in amperes is a) 2 b) 2 3 2 3 2 c) d) None of the above 4) Find the power dissipated by the 15 Ω resistor. a) 21.6 W b) 12 W c) 15.5 W d) None of the above 10 10 V 5) Find the Norton equivalent circuit between terminals a and b. a) Isc = 0.75 A, Req = 15 Ω b) Isc = 1.5 A, Req = 25 Ω c) Isc = 2 A, Req = 15 Ω 3A 10 d) Isc = 1 A, Req = 20 Ω 2A 15 10 V 10 10 a 20 6) Find the value of the inductor L, if vs(t) = 10cos10t V and the average power dissipated by the circuit is 4 W. a) 0.25 H b) 0.5 H c) 1 H d) 2 H b Problem 2: a) [4 points] For the circuit shown find the equivalent resistance seen between terminals a and b. SHOW ALL STEPS Step 8Ω || 8Ω = 4Ω 4Ω || 4Ω = 2Ω 2Ω || 2Ω = 1Ω 1Ω + 2Ω = 3Ω Req = 3 b) [6 points] Use node-voltage method to find the voltage v(t) in the circuit shown 4sin(4t ) V v(t) = Problem 3: [10 points] Find v(t) in the following circuit. 5cos(2t+30) V 1 3 v(t) 1 F 8 10cos(4t) A v(t) = Solution: Due to 5cos(2t+30) Due to 10cos(4t) Change to Phasor Domain Change to Phasor Domain +j2 3 V1 -j4 Combine two impedances in parallel Using VDR V1= [j4/(3+j4)]*5<30 Combine two impedances in parallel Using CDR I2=[3/(3-j4)]*10 = 3*10/5<-53.13 8V =4<90*5<30/5<53.13 = 4<66.87 I2= 6<+53.13 V2=I2*-j4= 24<-36.87 v1(t)=4cos(2t+66.87) v2(t)=24cos(4t-36.87) v(t) = v1(t) + v2(t) +v3(t) Due to 8V DC source = 4cos(2t+66.87)+24cos(4t+36.87)+8 3 v3(t) v3(t) = 8V 8V Problem 4: In the circuit shown, load 1 has impedance Z1 and load 2 has impedance Z2 Z1 absorbs 10 Watt (average power) and -15 var (Reactive power) Z2 absorbs 20 Watt (average power) and 30 var (Reactive power) 10 cos(300t − 45 ) A Find the following: a) [2 points] The apparent power delivered to load Z1 Answer = b) [2 points] The apparent power delivered to load Z2? Answer = c) [2 points] The power factor of load Z1 and indicate whether it is leading or lagging? pf1 = d) [2 points] The power factor of load Z2 and indicate whether it is leading or lagging? (e) [2 points] Find the voltage v(t)? a) b) v(t) = P1 = 10 − j15 watts P1 = 18.03 VA P1 = 20 + j 30 watts P1 = 36.06 VA c) pf1 = 10 /18.03 = 0.5546 leading d) pf1 = 20 / 36.06 = 0.5546 lagging e) Ptotal = 30 + j15 VA V = 2 Ptotal / I * = 2(30 + j15) /10∠45 = 6.7082∠ − 18.4349 = 6.3640 − j 2.1213 V v(t ) = 6.7082 cos(300t − 18.4349) V pf2 = Problem 5: [8 points] Three-phase power is supplied to a balanced, wye-connected load. The line-to-line voltage is 208 Vrms, and the load consumes a total power of 16 kwatt at a lagging power factor of 0.9. Determine: a) The magnitude of the transmission-line currents b) The individual load impedance ZL. The magnitude of the line current is= ZL = Solution: The transmission-line current (the load current) can be calculated as ( ) 16000 = 3 VL I L cos θ Z L ; IL = The phase voltage Vp=208/ 16000 = 49.346 A 3 (208)(0.9 ) =120 V. So, the magnitude of the individual load impedance can be calculated as ZL=Vp/IL=120/49.346=2.432 Ω. Since the power factor is lagging, θ Z = cos −1 (0.9 ) = 25.842 . Hence, ZL= 2.432∠25.842 = 2.189+j 1.061 Ω.