P28_070 - t = 0). Thus, with t = 0 . 00400 s, we obtain V =...

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70. In the steady state situation, the capacitor voltage will equal the voltage across the 15 kΩ resistor: V 0 =(15kΩ) µ 20 V 10 kΩ + 15 kΩ =12V . Now, multiplying Eq. 28-36 by the capacitance leads to V = V 0 e t/RC describing the voltage across the capacitor (and across the R = 15 kΩ resistor) after the switch is opened (at
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Unformatted text preview: t = 0). Thus, with t = 0 . 00400 s, we obtain V = (12) e − . 004 / (15000)(0 . 4 × 10 − 6 ) = 6 . 16 V . Therefore, using Ohm’s law, the current through the 15 kΩ resistor is 6 . 16 / 15000 = 4 . 11 × 10 − 4 A....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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