12V+ 8K + Io √xuk 2K m Vref=0 R K uk 2 Find I° by using nodal > 12:38 م 12:38 م
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- Convert the following instantaneous currents to phasors, using cos(t) as the reference. Give your answers in both rectangular and polar form. (a) i(t)=5002cos(t30) (b) i(t)=4sin(t+30) (c) i(t)=5cos(t15)+42sin(t+30)9. The matiueonk shawn is aparastineg the steady state with simusaielal un ualtage saunces. i v= 2cos 2t amd N2 = 2 sim 2+ , diturmine the valtage VaCt). 3173 F 2173 F Ja 2173F 31H allEvaluate 8/10° +62-20° 9/80° -4/50° 0.50 +j2.24 -0.50-j2.24 O 50-j224 O -50-j224 and express the result in rectangular form.
- Please show your detailed solution Number 14 The current in a series inductive circuit is 7.5 A at 25 Hz. The circuit takes 425 W, and the power factor is 0.47. The resistance of the circuit is A. 7.32 ohms C. 7.98 ohms B. 7.86 ohms D. 7.56 ohmsTwo impedances of Z = 100 +j50 Q and Z2 = 20 -j20 Q are connected in series. what is the magnitude of the equivalent impedance in 2? %3D %3D Your Answer: AnswerA non-sinusoidal current source (iT=2+1.5sin1000t+ 1cos3000t-0.5sin4000t) A is applied to the circuit shown. Find the total power? * 42 iL ww iT ic c -J92 J42 O 55 W 20 W 39 W O 28 W
- The type of the load whose current and voltage phasors are as shown in the phaser diagram is Im 90° - Re Pure capacitive load non-pure Inductive load Purse resistive load O Pure inductive load non-pure capacitive loadA non-sinusoidal current source (iT=2+1.5sin1000t+ 1cos30o0t- 0.5sin4000t) A is applied to the * ?circuit shown. Find the total power İL iT ic -J9N J4Q 39 W O 20 W O 55 W 28 W )!( هذا السؤال مطلوبSolve for x & y. No solution, final answers only in rectangular form. x + y = 520° – 3x + 2y = 3+j4
- Use nodal analysis to find Io. -j2N 6290°v j5N 220° AÀ -j1QDC-AC Circuits Express R1 using Rx, Ry, and Rz when performing wye to delta conversion R3 Rx Rz R2 R1 Ry O A. (Rx+Ry)(Ry+Rz)(Rx+Rz) Rx В. RyRz Rx+Ry+Rz O C. Ry+Rz RXRYRZ O D. RXRYRZ Ry+Rz OE. RxRy+RyRz+RXRZ RxSolve the power triangle at Z6 in the circuit diagram below.V1, V2, and V3 = 220∠0°, 50 HzZ1 = 7+j8 ohmsZ2 = 6+j3 ohmsZ3 = 5+j5 ohmsZ4 = 2+j3 ohmsZ5 = 8+j4 ohmsZ6 = 1+j9 ohms