142_pdfsam_math 54 differential equation solutions odd

142_pdfsam_math 54 differential equation solutions odd -...

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Chapter 3 3. In this RC circuit, R = 100 Ω, C =10 12 F, the initial charge of the capacitor is Q = q (0) = 0 coulombs, and the applied constant voltage is V = 5 volts. Thus we can use a general equation for the charge q ( t ) of the capacitor derived in Example 2. Substitution of given data yields q ( t )= CV +[ Q CV ] e t/RC =10 12 (5) ± 1 e t/ ( 100 · 10 12 ) ² =5 · 10 12 ± 1 e 10 10 t ² and so E C ( t )= q ( t ) C =5 ± 1 e 10 10 t ² . Solving the equation E C ( t ) = 3, we get 5 ± 1 e 10 10 t ² =3 e 10 10 t =0 . 4 t = ln 0 . 4 10 10 9 . 2 × 10 11 (sec) . Therefore, it will take about 9 . 2 × 10 11 seconds for the voltage to reach 3 volts at the receiving gate. 5. Let V ( t ) denote the voltage across an element, and let I ( t ) be the current through this element.
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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