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891_PartUniversity Physics Solution

# 891_PartUniversity Physics Solution - Inductance(b i =...

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Inductance 30-15 (b) 2 ( / ) 2 2( / ) R L t R L t L di R dU di i e i iL Ri e R dt L dt dt R = = − = = − = E E . 4 2 2(240/0.160)(4.00 10 ) (60 V) 4.52 W. 240 L dU e dt × = − = − Ω (c) In the resistor, 4 2 2 2 2( / ) 2(240/0.160)(4.00 10 ) (60 V) 4.52 W 240 R L t R R dU P i R e e dt R × = = = = = Ω E . (d) 2 2 2( / ) ( ) R L t R P t i R e R = = E . 2 2 2 2( / ) 3 2 0 (60 V) (0.160 H) 5.00 10 J, 2 2(240 ) R L t R L U e R R R = = = = × Ω E E which is the same as part (a). E VALUATE : During the decay of the current all the electrical energy originally stored in the inductor is dissipated in the resistor. 30.55. I DENTIFY and S ET U P : Follow the procedure specified in the problem. 2 1 2 Li is the energy stored in the inductor and 2 /2 q C is the energy stored in the capacitor. The equation is 0. di q iR L dt C = E XECUTE : Multiplying by ° i gives 2 0. di qi i R Li dt C + + = ( ) ( ) 2 2 1 1 1 2 2 2 2 , L d d d di di U Li L i L i Li dt dt dt dt dt = = = = the second term. 2 2 1 1 ( ) (2 ) , 2 2 2 C d d q d dq qi U q q dt dt C C dt C dt C = = = = the third term. 2 , R i R P = the rate at which electrical energy is dissipated in the resistance. , L L d U P dt = the rate at which the amount of energy stored in the inductor is changing. , C C d U P dt = the rate at which the amount of energy stored in the capacitor is changing. E VALUATE : The equation says that 0; R L C P P P + + = the net rate of change of energy in the circuit is zero. Note that at any given time one of or C L P P is negative. If the current and L U are increasing the charge on the capacitor and C U are decreasing, and vice versa. 30.56. I DENTIFY : The energy stored in a capacitor is 2 1 2 C U Cv = . The energy stored in an inductor is 2 1 2 L U Li = . Energy conservation requires that the total stored energy be constant. S ET U P : The current is a maximum when the charge on the capacitor is zero and the energy stored in the capacitor is zero. E XECUTE : (a) Initially 16.0 V v = and 0 i = . 0 L U = and 2 6 2 4 1 1 2 2 (5.00 10 F)(16.0 V) 6.40 10 J C U Cv = = × = × . The total energy stored is 0.640 mJ . (b) The current is maximum when 0 q = and 0 C U = . 4 6.40 10 J C L U U + = × so 4 6.40 10 J L U = × . 2 4 1 max 2 6.40 10 J Li = × and 4 max 3 2(6.40 10 J) 0.584 A 3.75 10 H i × = = × . E VALUATE : The maximum charge on the capacitor is 80.0 C Q CV μ = = . 30.57. I DENTIFY and S ET U P : Use 2 1 2 C C U CV = (energy stored in a capacitor) to solve for C . Then use Eq.(30.22) and 2 f ω π = to solve for the L that gives the desired current oscillation frequency. E XECUTE : 2 2 2 1 2 12.0 V; so 2 / 2(0.0160 J)/(12.0 V) 222 F C C C C C V U CV C U V μ = = = = = 2 1 1 so (2 ) 2 f L f C LC π π = = 3500 Hz gives 9.31 H f L μ = = E VALUATE : f is in Hz and ω is in rad/s; we must be careful not to confuse the two. 30.58. I DENTIFY : Apply energy conservation to the circuit. S ET U P : For a capacitor / V q C = and 2 /2 U q C = . For an inductor 2 1 2 U Li = E XECUTE : (a) 6 max 4 6.00 10 C 0.0240 V.

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891_PartUniversity Physics Solution - Inductance(b i =...

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