143_pdfsam_math 54 differential equation solutions odd

143_pdfsam_math 54 differential equation solutions odd - e...

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Exercises 3.6 (c) Capacitor . Here, with q ( t ) denoting the electrical charge on the capacitor, V ( t )= E C ( t )= 1 C q ( t ) q ( t )= CE C ( t ) I ( t )= dq ( t ) dt = d [ CE C ( t )] dt and so P C = d [ CE C ( t )] dt E C ( t )= C 2 ± 2 E C ( t ) dE C ( t ) dt ² = C 2 d [ E C ( t ) 2 ] dt = d [ CE C ( t ) 2 / 2] dt . 7. First, we fnd a ±ormula ±or the current I ( t ). Given that R =3Ω , L = 10 H, and the voltage ±unction E ( t )isaconstant ,say , V , the ±ormula (3) on page 121 (which describes currents in RL circuits) becomes I ( t )= e 3 t/ 10 ³Z e 3 t/ 10 V 10 dt + K ´ = e 3 t/ 10 ³ V 3 e 3 t/ 10 + K ´ = V 3 + Ke 3 t/ 10 . The initial condition, I (0) = 0 (there were no current in the electromagnet be±ore the voltage source was applied), yields 0= V 3 + Ke 3(0) / 10 K = V 3 I ( t )= V 3 ( 1 e 3 t/ 10 ) . Next, we fnd the limiting value I I ( t ), that is, I = lim t →∞ ± V
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Unformatted text preview: e 3 t/ 10 ) = V 3 (1 0) = V 3 . Thereore, we are looking or the moment t when I ( t ) = (0 . 9) I = (0 . 9) V/ 3. Solving yields . 9 V 3 = V 3 ( 1 e 3 t/ 10 ) e 3 t/ 10 = 0 . 1 t = 10 ln 0 . 1 3 7 . 68 . Thus it takes approximately 7 . 68 seconds or the electromagnet to reach 90% o its fnal value. EXERCISES 3.6: Improved Eulers Method, page 132 1. Given the step size h and considering equally spaced points we have x n +1 = x n + nh, n = 0 , 1 , 2 , . . . . 139...
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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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