Find the Thévenin equivalent network seen by the capacitor in Figure P5.30 for
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Principles and Applications of Electrical Engineering
- 2 At t < 0, the circuit shown in Figure P5.22 is at steady state. The switch is changed as shown at t = 0. Vsi = 35 V C = 11 µF Vsz = 130 V R = 17 k2 R2 = 7 k2 R = 23 k2 Determine at t = 0+ the initial current through R just after the switch is changed. 1= 0 R3 Vs1 Vs2arrow_forward7 Find the voltage across C in the circuit of Figure P5.57 for t> 0. Let G = 5 µF; C = 10 µF. Assume the capacitors are initially uncharged. 12 19 2 ww t= 0 10 v(* C2arrow_forward3. An RC circuit has an emf given by 400 cos2t volts, a resistance of 100 ohms and a capacitance of 0.01 Farad. Initially there is no charge on the capacitor. Find the current in the circuit after 0.5 second.arrow_forward
- Consider the R-C circuit. we idealise the emf to be constant and have zero internal resistance. We begin with capacitor initially uncharged. At initial time t=0, the switch was closed. Answer the questions attached.arrow_forward1 Just before the switch is opened at t = 0 in Figure P5.21, the current through the inductor is 1.70 mA in the direction shown. Vs = 12 V L = 0.9 mH R = 6 k2 R2 = 6 k2 R = 3 k2 Determine the time constant of the circuit for t > 0.arrow_forwardA circuit is designed with an AC source of max voltage 12 and frequency 60 Hz. The circuit has a resistance of 1050 Ohms, an inductance of 0.06 Henrys, and a capacitance of 0.009 coulombs per volt. - omega for source in rad/s? - omegaR for circuit? -XL? -XC? -phi in radians? -Z? -imax?arrow_forward
- .A series RLC circuit contains a 4-kN resistor, an inductor with an inductive reactance (X,) of 3.5 kn, and a capacitor with a capacitive reactance (Xc) of 2.4 kN. A 120-Vac, 60-Hz power source is connected to the circuit. How much voltage is dropped across the inductor?arrow_forwardSolve for the node voltages shown in Figure P5.53. 10/0 (+ 10 2 +j20 2 15 n 1 Figure P5.53 000arrow_forwardSolve the circuit by obtaining the state equation. The initial condition voltage value of the capacitance element is VC (0) = 1Volt. R1 = R2 = R3 = R4 = 1Ω, E = 3Volt. State the core solution and the forced solution components in the solution you obtained.arrow_forward
- 4 If the switch in the circuit shown in Figure P5.64 is closed at t = 0 and Vs = 12 V C = 130 µF R = 2.3 k2 R, = 7 k2 L= 30 mH determine the current through the inductor and the voltage across the capacitor and across Rị after the circuit has returned to a steady state. t= 0 R1 Vs R2arrow_forwardThe time taken by the series RL circuit having an inductance of 0.6 H and resistance of 30 Ohms to reach a steady-state value.arrow_forward7 Steady-state conditions exist in the circuit shown in Figure P5.27 at t < 0. The switch is closed at t = 0. V = 12 V R = 0.68 k2 R = 2.2 k2 R = 1.8 k2 C= 0.47 µF Determine the current through the capacitor at t = 0+, just after the switch is closed. ww. idt) R. t= 0 R1 Ry ww-arrow_forward
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