# hw18 - Physics 21 Fall 2009 Solution to HW-18 30.19 An...

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Unformatted text preview: Physics 21 Fall, 2009 Solution to HW-18 30.19 An inductor with an inductance of 2.50 H and a resistance of 8 . 00 Ω is connected to the terminals of a battery with an emf of 6.00 V and negligible internal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current is 0.500 A; (c) the current 0.250 s after the circuit is closed; (d) the final steady-state current. (a) Applying Kirchhoff’s Law, we find the loop equation is E − iR − L di dt = 0 ⇒ di dt = E − iR L . The initial condition is that i = 0 at t = 0, so at that time di dt = E L = 6 V 2 . 5 H = 2 . 4 A / s . (b) Using the loop equation again for i = 0 . 5 A , di dt = E − iR L = 6 V − (0 . 5 A × 8 Ω) 2 . 5 H = 0 . 8 A / s . (c) Using the general solution for the loop equation, we have i = E R (1 − e- ( R/L ) t ) = 6 V 8 Ω ³ 1 − e- (8Ω / 2 . 5 H)0 . 25 s ´ = 0 . 413 A . (d) The general solution shows that as t → ∞ , the exponen- tial term vanishes. Then i = E R = 6 V 8 Ω = 0 . 75 A ....
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hw18 - Physics 21 Fall 2009 Solution to HW-18 30.19 An...

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