896_PartUniversity Physics Solution

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

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30-20 Chapter 30 30.68. IDENTIFY: Apply the loop rule to the two loops. The current through the inductor doesn’t change abruptly. SET UP: For the inductor di L dt = E and E is directed to oppose the change in current. EXECUTE: (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 vL dt dt =⇒ = = = The top circuit loop: 60.0 11 1 60.0 V V1 . 5 0 A . 40.0 iR i = = Ω The bottom loop: 22 2 45.0 V 60 V 15.0 V 0 1.80 A. 25.0 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 12 2.40 A. ii EVALUATE: The current through 1 R changes abruptly when the switch is closed. 30.69. IDENTIFY and SET UP: 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 EXECUTE: (a) At time 00 0 0, 0 so 0. ac ti v i R By the loop rule 0, ac cb vv −= E so 36.0 V. cb ac =− == EE ( 0 0 = 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 . iR R −+ = E 0 0 36.0 V 0.180 A 50.0 150 i RR = + Ω E (0.180 A)(50.0 ) 9.0 V ac vi R Ω = Thus 0 0 (0.180 A)(150 ) 0 27.0 V cb di R L dt =+ = Ω + = (Note that . ac cb + = E ) (c) 0 ac cb −−= E 0 0 di iR iR L dt −−− = E 0 a n d di L di Li R R i dt R R dt R R ⎛⎞ =− + = ⎜⎟ ++ ⎝⎠ E E 0 0 /( ) di R R dt R L + = + E Integrate from t = 0, when i = 0, to t , when 0 : = 0 0 0 ln , /( ) i it di R R R R dt i t R L R R L ⎡⎤ + = ⎢⎥ + + ⎣⎦ ∫∫ E E so 0 0 ln ln L + 0 0 /( ) ln /( ) R R R t L + + + E E Taking exponentials of both sides gives 0 / 0 /( ) /( ) RRtL R e = + E E and 0 / 0 0 1 ie + 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 tt e −Ω = Ω+ Ω
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Inductance 30-21 At 0 0, (0.180 A)(1 1) 0 ti →= = (agrees with part (a)). At 0 , (0.180 A)(1 0) 0.180 A →∞ = = (agrees with part (b)). () 0 / /0.020 s 0 00 0 1 9.0 V 1 RRtL t ac R vi R e e RR −+ == = + E /0.020 s /0.020 s 36.0 V 9.0 V(1 ) 9.0 V(3.00 ) tt cb ac vv e e −− =− = = + E At 0, 0, 36.0 V ac cb tv v →== (agrees with part (a)). At 9.0 V, 27.0 V ac cb v = = (agrees with part (b)). The graphs are given in Figure 30.69b. Figure 30.69b EVALUATE: The expression for i ( t ) we derived becomes Eq.(30.14) if the two resistors 0 R and R in series are replaced by a single equivalent resistance 0 . R R + 30.70. IDENTIFY: Apply the loop rule. The current through the inductor doesn’t change abruptly. SET UP: With 2 S closed, cb v must be zero.
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896_PartUniversity Physics Solution - 30-20 Chapter 30...

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