Fin d Vi , V2 3,-Vz =6--0 %3D -Vi+3½=12 --- 3/2
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- 7. The resistor in an R-L series circuit has a voltage drop of 53 V, and the inductor has a voltage drop of 28 V. What is the applied voltage of the circuit?The circuit in Figure 24-2 is connected to a 120-V, 60-Hz line. The resistor has a resistance of 36 , the inductor has an inductive reactance of 40 , and the capacitor has a capacitive reactance of 50 . ET120VITZVAPFERIRR36PELILXL40VARsLLECICXC50VARsCCThe circuit in Figure 24-2 is connected to a 1000-Hz line. The resistor has a current flow of 60 A, the inductor has a current flow of 150 A, and the capacitor has a current flow of 70 A. The circuit has a total impedance of 4.8 . ETITZ4.8VAPFERIR60ARPELIL150AXLVARsLLECIC150AXCVARsCC
- An R-L series circuit contains two resistors and two inductors. The resistors are 86 k and 68 k . The inductors have inductive reactances of 24 k and 56 k . The total voltage is 480 volts. Find the voltage drop across the 56-k inductor.The circuit shown in Figure 24-2 has a current of 38 A flowing through the resistor, 22 A flowing through the inductor, and 7 A flowing through the capacitor. What is the total circuit current?5. The hypotenuse has a length of 65 in., and side A has a length of 31 in. What is angle X?
- Two series capacitors C1 and C2 are connected in series to a 24V battery. The dielectric for between the plates are K1=3.0 and K2= 4.0. Without any dielectric, capacitors C1 and C2 had equal capacitance of 5uF. What is the voltage across C2?What is the solution for SΔgen= 3Va’’ ()= –5.1– j4.7VA Where Va’’=Volts Va’=0.9∠-10.9VoltsFor the circuit shown below calculate IE in mA when IDSS=12mA VP=-6V and B=120