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896_PartUniversity Physics Solution

# 896_PartUniversity Physics Solution - 30-20 Chapter 30...

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30-20 Chapter 30 30.68. I DENTIFY : Apply the loop rule to the two loops. The current through the inductor doesn’t change abruptly. S ET U P : For the inductor di L dt = E and E is directed to oppose the change in current. E XECUTE : (a) Switch is closed, then at some later time 50.0 A/s (0.300 H) (50.0 A/s) 15.0 V. cd di di v L dt dt = = = = The top circuit loop: 60.0 1 1 1 60.0 V V 1.50 A. 40.0 i R i = = = Ω The bottom loop: 2 2 2 45.0 V 60 V 15.0 V 0 1.80 A. 25.0 i R i = = = Ω (b) After a long time: 2 60.0 V 2.40 A, 25.0 i = = Ω and immediately when the switch is opened, the inductor maintains this current, so 1 2 2.40 A. i i = = E VALUATE : The current through 1 R changes abruptly when the switch is closed. 30.69. I DENTIFY and S ET U P : The circuit is sketched in Figure 30.69a. Apply the loop rule. Just after 1 S is closed, i = 0. After a long time i has reached its final value and di / dt = 0. The voltage across a resistor depends on i and the voltage across an inductor depends on di / dt. Figure 30.69a E XECUTE : (a) At time 0 0 0 0, 0 so 0. ac t i v i R = = = = By the loop rule 0, ac cb v v = E so 36.0 V. cb ac v v = = = E E ( 0 0 i R = so this potential difference of 36.0 V is across the inductor and is an induced emf produced by the changing current.) (b) After a long time 0 0 di dt so the potential 0 di L dt across the inductor becomes zero. The loop rule gives 0 0 ( ) 0. i R R + = E 0 0 36.0 V 0.180 A 50.0 150 i R R = = = + Ω + Ω E 0 0 (0.180 A)(50.0 ) 9.0 V ac v i R = = Ω = Thus 0 0 (0.180 A)(150 ) 0 27.0 V cb di v i R L dt = + = Ω + = (Note that . ac cb v v + = E ) (c) 0 ac cb v v = E 0 0 di iR iR L dt = E 0 0 0 ( ) and di L di L i R R i dt R R dt R R = + = − + + + E E 0 0 /( ) di R R dt i R R L + = − + + E Integrate from t = 0, when i = 0, to t , when 0 : i i = 0 0 0 0 0 0 0 0 0 ln , /( ) i i t di R R R R dt i t i R R L R R L + + = = − − + = − + + + E E so 0 0 0 0 ln ln R R i t R R R R L + + = − + + E E 0 0 0 0 /( ) ln /( ) i R R R R t R R L + + + = − + E E Taking exponentials of both sides gives 0 ( ) / 0 0 0 /( ) /( ) R R t L i R R e R R + + + = + E E and ( ) 0 ( ) / 0 0 1 R R t L i e R R + = + E Substituting in the numerical values gives ( ) ( ) (200 / 4.00 H) /0.020 s 0 36.0 V 1 (0.180 A) 1 50 150 t t i e e Ω = = Ω + Ω

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Inductance 30-21 At 0 0, (0.180 A)(1 1) 0 t i = = (agrees with part (a)). At 0 , (0.180 A)(1 0) 0.180 A t i → ∞ = = (agrees with part (b)). ( ) ( ) 0 ( ) / /0.020 s 0 0 0 0 1 9.0 V 1 R R t L t ac R v i R e e R R + = = = + E /0.020 s /0.020 s 36.0 V 9.0 V(1 ) 9.0 V(3.00 ) t t cb ac v v e e = = = + E At 0, 0, 36.0 V ac cb t v v = = (agrees with part (a)). At , 9.0 V, 27.0 V ac cb t v v → ∞ = = (agrees with part (b)). The graphs are given in Figure 30.69b.
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896_PartUniversity Physics Solution - 30-20 Chapter 30...

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