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?
5.137 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

. 01.   VC = (V0)*{1 -exp[(-t)/()]}      ;   equationfor the voltage on a charging capacitor in a RC circuit . 02.   = 9.10E-3seconds      ;   thisis derived from the data in the problem statement... (1 - 1/e) is9.10 milliseconds. 03.   V0 = 6volts      ;   Givenin the problem statement 04.   VC = (6 volts)*{1- exp[(-17.65E-3)/(9.10E-3)] } = 5.137397219 volts =5.137 volts 05.   =R*C         C= /R 06.   C = (9.10E-3 seconds)/(82.0 ) =110.9756E-6 Farads ˜ 111F .

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