2.) Consider Activity 2-2. Initially, the switch is in the“ 2 ” position and has

Question

2.) Consider Activity 2-2. Initially, the switch is in the“2” position and has been there for avery long time (“very longtime” meaning many, many time constants). The switch is movedto the “1” position and the voltageacross the capacitor reaches“1 - 1/e” (about 63%) of the nominal6 V after 9.10 ms.

a) What is the voltage across the capacitor 17.65 ms after the switch is moved to the“1" position?
7 V

You remove the “100  ” resistorfrom the circuit and find that is has a resistance of 82.0 .

b) What is C?
8 µF

Explanation / Answer

Normal voltage V = 6 Volt time constant T = 9.10 ms = 9.1 * 10 ^ -3 s (a). time t = 17.65 ms = 17.65 * 10 ^ -3 s Voltage after time t is V ' = V [ 1- e-t / T ]                                         = 6 [ 1- e - 1.939 ]                                          = 6[ 1- 0.1437 ]                                         = 5.137 Volts

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.