A long time after the switch has been closed on the circuit shown in the figure...
a. What is the output voltage Vout? Now the switch is opened and the output voltage increases. When the output voltage reaches 10.0 V, the switch closes and then the capacitor begins to discharge. When the output voltage reaches 5.0 V, the switch opens again. Thus this circuit undergoes periodic cycles of charging and discharging.b. What is the time constant during a charging phase?c. What is the time constant during a discharging phase?d. What length of time passes between subsequent 10.0 V output voltages?e. If...
Hello, Can anyone please solve part c of this question and give me the final answer ? I tried to solve it, but it doesn't work ...In the circuit of the figure below, the switch S has been open for a long time. It is then suddenly closed. Take ε=10.0 V, R₁=49.0 k Ω, R₂=130 k Ω, and C=15.0 μ F.