Use nodal method to find the branch currents l1, l2, 13, l4 and Is in the network shown below.
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please don't round off the calculated values
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- -The Vollage V= 100210°V is 62 pplied across a capacilive 1oad . The 1-a kes ap paraal' Power Load of 500VA teading Power fracl-or. The reaclance (xc) of load 's---- 7What's the rise time for this system with its closed-loop poles as shown? 14 -6 -4 12 -j6 Select one: a. 0.607 sec b. 3.14 sec C. 1.80 sec d. 0.254 seC e. 0.554 secIn the circuit below the currents are named I, la, and le The current direction is determined by the source (out of positive terminal) in the middle and right branches and is clockwise in the left branch. • la flows in the left branch of the circuit • Is flows in the middle branch of the circuit • Icflows in the right branch of the circuit АВ A B R2 R3 N When applying KVL: R4 • transverse the loop in a Clockwise direction • voltage drops are negative. R1 Vc Vb label the voltage across the resistors with the same numeric subscript as the resistor R1:= 1.2kN R2= 1.8kN R3:= 1.5kN R4= 0.5kN V= 10V Ve= 6V
- Qs # 2. Write down the equations that results from applving Kirchhoff's junction and loop rule. A Rz 1, 7V 3v 5VSolve 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 ohmsThree impedances, Z1 =10020.Q, Z2 =63.2518.43.2 and Z3 =1002-90•Q are connected in star. Convert the star to an equivalent delta connection. And draw the star network and also the equivalent of delta connection
- DC-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 for the node voltage V₁ -jsn So 20190 102 www +100 12 10/04. Solve for I, I2, Iz and I4 in the ckt below using Cramer's Rule. ISMAT169NUMI 252 METHO 169NUMERICALIM SANDANALYSISM THOANDANALY RICLMET 169NUICA MAT169NUMERICA 幸 DDSANDA MAT169NUMERI THODSANDANALYSSMAT ANALYSISMAT 9NUM METHODSA 12V (1) I2 1 YSISMAT169N CALMETHODSANDA 351 ETHO SISMATI6 OSANDANAL RICALMETHODSAND NDAN SAND ANALYSISMAT 丰 RICAM MERICALME NDANALYSISM NUMERICAL METHOSAN SAND ALYSI METH JALYS 69NUMERI ISN I) I3 ERICALM ooDANAL TRICALMET NDANASA SMAT169NUAALYSISMAT169NUMERICALMETHODSA AT 169NU SANDA MERICLMETHODSANDA
- For the circuit of the following figure, if R=14Q, Zc=-J26 Q, and ZL=J10 (Q, find the value of the current through ZL. Zc V1 1.5 /24° V ZL -) 3-41° V Oa. 173.759 L -81.69 ° mA Ob. 175.759L-71.69 ° mA Oc. 175.759L-81.69 ° mA Od. 173.759 L -71.69 ° mASolve the power triangle at Z6 in the circuit diagram below.V1, V2, and V3 = 220∠0° V, 50 HzZ1 = 7+j8 ohmsZ2 = 6+j3 ohmsZ3 = 5+j5 ohmsZ4 = 2+j3 ohmsZ5 = 8+j4 ohmsZ6 = 1+j9 ohmsNeed to use kirchhoffs voltage and circuit laws. But not understanding. Need to find V500,V50,V150,V250,V1000,P1000.