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chapter 27

# chapter 27 - Problem 12 The current in an inductor is...

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Problem 12. The current in an inductor is changing at the rate of 100 A/s, and the inductor emf is 40 V. What is its self-inductance? Solution From Equation 32-5, L = E = ( dI = dt ) = 40 V = ( 100 A / s ) = 0 . 4 H . Problem 15. A cardboard tube measures 15 cm long by 2.2 cm in diameter. How many turns of wire must be wound on the full length of the tube to make a 5.8-mH inductor? Solution From Equation 32-4, N = l = μ 0 A = [( 5 . 8 mH )( 15 cm ) = ( 4 π × 10 7 H / m ) ( 1 . 1 cm ) 2 ] 1 = 2 = 1 . 35 × 10 3 turns. Problem 16. The current in a 2.0-H inductor is given by I = 3 t 2 + 15 t + 8, where t is in seconds and I in amperes. Find an expression for the magnitude of the inductor emf. Solution From Equation 32-5, E = L dI = dt = 2 ( 6 t + 15 ), where E is in volts and t is in seconds. Problem 18. The current in a 40-mH inductor is given by I = I 0 e bt , where I 0 = 10 A and b = 20 s 1 . What is the magnitude of the inductor emf at (a) t = 0, (b) t = 25 ms, and (c) t = 50 ms? Solution The magnitude of the inductor emf is E = L dI = dt = LI 0 be bt , from Equation 32-5. (a) At t = E ( 0 ) = ( 40 mH ) × (10 A)(20 s 1 ) = 8 V, so (b) at t = 25 ms , E ( t ) = ( 8 V ) e 0 . 5 = 4 . 85 V, and (c) E ( 50 ms ) = 2 . 94 V . Problem 21. The emf in a 50-mH inductor is given by E = E p sin ω t , where E p = 75 V and = 140 s 1 . What is the peak current in the inductor? (Assume the current swings symmetrically about zero.) Solution From Equation 32-5, dI = dt = ( E p = L ) sin t , so integration yields I ( t ) = ( E p = L ) cos t . (Since I ( t ) is symmetric about I = the constant of integration is zero.) The peak current is I p = E p = L = 75 V = ( 140 s 1 × 50 mH ) = 10 . 7 A . Problem 4. Two coils have a mutual inductance of 580 mH. One coil is supplied with a current given by I = 3 t 2 2 t + 4, where I is in amperes and t in seconds. What is the induced emf in the other coil at time t = 2.5 s?

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