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Principles and Applications of Electrical Engineering
- =7 Determine the current through the capacitor just before and just after the switch is closed in Figure P5.37. Assume steady-state conditions for t < 0. V = 12 V C = 150 µF R = 400 m2 R2 = 2.2 k2 t = 0 R1 + SI +)arrow_forwardB Determine the voltage across the inductor just before and just after the switch is changed in Figure P5.38. Assume steady-state conditions exist for t < 0. Vs = 12 V Rs = 0.24 2 R = 33 k2 L = 100 mH t= 0 Rs + EIarrow_forward2 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_forward
- 8 For 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 R = 7 k2 R = 23 k2 Determine the time constant of the circuit for t> 0.arrow_forwardGiven circuit below, use superposition to find voltage across the capacitor, vclt). Frequency is 100 Hz. 6kn 4kn reee zkn O SmA <45 Vc (t) DC a) Given circuit below and switch ciosed for long time, what is the value of Vc? 5mA 3 luk bị At0, switch is opened. Write a mathematical expression for Velt) after opening of the switch. Evaluate this voltage at te10 ms. Attach File Browse Local Fies rowie Conent Cotection 74°Farrow_forwardFor the network shown in Figure (Q5), Find: 1. The total equivalent inductance (Lequiv.). 2. If the current waveform shown in Figure (6) is applied to the (Leguiv). Determine an expression for the voltage across it. 6H 10 H 9H 4H SH 3 H 20 H 13H Lequiv. Figure (Q5) i (A) - (ms) 10 12 -5- Figure (6)arrow_forward
- 6 At t< 0, the circuit shown in Figure P5.66 is at steady state, and the voltage across the capacitor is +7 V. The switch is changed as shown at t= 0, and Vs = 12 V C= 3,300 µF R = 9.1 k2 R = 4.3 k2 R3 = 4.3 k2 L= 16 mH Determine the initial voltage across R2 just after the switch is changed. t=0 Le )V½ R R 2 R3 ww-arrow_forwardAt 0-, no currrent flows through the capacitors because they are open, how did you combine the capacitors for the voltage divider since it is the capacitance value and not reactance, is it right? Can we just combine the capacitance value? Please explain. I did not understand..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
- The 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_forward4 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_forward5. RC Charging Circuit is shown in the figure. Find the differantial equation and its solution of represent the charcing circuit using KIRCHHOFF's RULE by meaning changing the t- 0 R= 47kn charges with respect to time. (Vc =%, 1 = ) and then Calculate: dt a. q(t = 100) =? The charge across the Capacitor after 100 seconds b. Vc(t = 100) =?The voltage across the Capacitor after 100 seconds E=5V Vc C- 1000uF l1=10 Aarrow_forward
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