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 R = 10.0 kΩ and C = 10.0 μF, what is the frequency of operation? (Note: This circuit is a simplified example of a timer circuit, a crucial element in many electronics applications.)
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In the circuit shown in (Figure 1), the switch S has been closed for a long time.
The switch in the circuit below has been closed for a very long time. 2. a. What is the voltage across the capacitor? b. If the switch is opened, what is the time constant for discharging the capacitor? c. how long does it take the capacitor to discharge to 1/10th of its initial voltage? R1 = 1.00 Ω R2 = 8.00 Ω R3 = 4.00 Ω R4 = 2.00 Ω C = 1.00 μF Battery = 10.0 V R2 R4
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