881_PartUniversity Physics Solution

881_PartUniversity Physics Solution - Inductance 30.20....

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Inductance 30-5 30.20. IDENTIFY: The current decays exponentially. SET UP: After opening the switch, the current is / 0 , tR L iI e = and the time constant is τ = L / R . EXECUTE: (a) The initial current is 0 I = (6.30 V)/(15.0 ) = 0.420 A. Now solve for L and put in the numbers. 0 (2.00 ms)(15.0 ) 43.3 mH 0.210 A ln( / ) ln 0.420 A tR L −− Ω == = ⎛⎞ ⎜⎟ ⎝⎠ (b) = L/R = (43.3 mH)/(15.0 ) = 2.89 ms (c) Solve / 0 t e = for t , giving 0 ln( / ) (2.89 ms)ln(0.0100) 13.3 ms. ti I =− = EVALUATE: In less than 5 time constants, the current is only 1% of its initial value. 30.21. IDENTIFY: / /( 1 ) , t iR e E with / . L R = The energy stored in the inductor is 2 1 2 . UL i = SET UP: The maximum current occurs after a long time and is equal to / . R E EXECUTE: (a) max / = E so max /2 ii = when 1 2 (1 ) t/ τ e = and 1 2 . t/ τ e = ( ) 1 2 /l n. t τ −= 3 ln 2 (ln 2)(1.25 10 H) 17.3 s 50.0 L t μ R × = Ω (b) 1 max max 2 when 2. UU i i / 11 2 , t τ e so / e 1 1 2 0.2929. t τ = ln(0.2929)/ 30.7 s. tL R μ = EVALUATE: 5 / 2.50 10 s 25.0 s. LR τμ ==× = The time in part (a) is 0.692 and the time in part (b) is 1.23 . 30.22. IDENTIFY: With 1 S closed and 2 S open, ( ) it is given by Eq.(30.14). With 1 S open and 2 S closed, ( ) is given by Eq.(30.18). SET UP: 2 1 2 . i = After 1 S has been closed a long time, i has reached its final value of / . IR = E EXECUTE: (a) 2 1 2 I = and 2 2(0.260 J) 2.13 A. 0.115 H U I L = (2.13 A)(120 ) 256 V. IR = = E (b) (/) RLt e = and ( ) 22 2 ( / ) 2 1 1 1 0 2 2 2 . RL t i L I e U L I = = 2( / ) 1 2 , e = so () 4 0.115 H ln ln 3.32 10 s. 2 2(120 ) L t R = × Ω EVALUATE: 4 /9 . 5 8 1 0 s . The time in part (b) is ln(2)/2 0.347 . = 30.23. IDENTIFY: L has units of H and R has units of Ω . SET UP: 1 H 1 s EXECUTE: Units of / H/ ( s)/ s = Ω ⋅Ω = = units of time. EVALUATE: // RtL t = is dimensionless. 30.24. IDENTIFY: Apply the loop rule. SET UP: In applying the loop rule, go around the circuit in the direction of the current. The voltage across the inductor is / . Ldi dt EXECUTE: /0 . Ldi dt iR = di R i dt L gives 0 0 I di R dt iL and 0 ln( / ) . R t L 0 . tRL e = EVALUATE: / di dt is negative, so there is a potential rise across the inductor; point c is at higher potential than point b . There is a potential drop across the resistor. 30.25. IDENTIFY: Apply the concepts of current decay in an R - L circuit. Apply the loop rule to the circuit. i ( t ) is given by Eq.(30.18). The voltage across the resistor depends on i and the voltage across the inductor depends on di / dt . SET UP: The circuit with 1 S closed and 2 S open is sketched in Figure 30.25a. 0 di iR L dt E Figure 30.25a Constant current established means 0. di dt = EXECUTE: 60.0 V 0.250 A 240 i R = Ω E
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30-6 Chapter 30 (a) SET UP: The circuit with 2 S closed and 1 S open is shown in Figure 30.25b. (/) 0 RLt iI e = At 0 0, 0.250 A ti I == = Figure 30.25b The inductor prevents an instantaneous change in the current; the current in the inductor just after 2 S is closed and 1 S is opened equals the current in the inductor just before this is done. (b) EXECUTE: 4 ( / ) (240 /0.160 H)(4.00 10 s) 0.600 0 (0.250 A) (0.250 A) 0.137 A e e e −− Ω ×− = = (c) SET UP: See Figure 30.25c.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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881_PartUniversity Physics Solution - Inductance 30.20....

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