891_PartUniversity Physics Solution

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

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Inductance 30-15 (b) 2 (/) 2 2 RLt L di R dU di ie i i L R i e Rd t L d t d tR = =− = = EE . 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 22 2 2( / ) 2(240/0.160)(4.00 10 ) (60 V) 4.52 W 240 R R dU Pi R e e dt R −− × == = = = Ω E . (d) 2 ( / ) () R Pt iR e R E . 2 2( / ) 3 2 0 (60 V) (0.160 H) 5.00 10 J, 2 2(240 ) R L Ue RR R = = × Ω which is the same as part (a). EVALUATE: During the decay of the current all the electrical energy originally stored in the inductor is dissipated in the resistor. 30.55. IDENTIFY and SET UP: Follow the procedure specified in the problem. 2 1 2 Li is the energy stored in the inductor and 2 /2 qC is the energy stored in the capacitor. The equation is 0. di q iR L dt C = EXECUTE: Multiplying by & i gives 2 0. di qi iR L i dt C + += 11 1 2 2, L dd d d id i UL i L i L i L i dt dt dt dt dt ⎛⎞ === = ⎜⎟ ⎝⎠ the second term. 2 2 ( 2 ) , 2 C d d q d dq qi Uq q dt dt C C dt C dt C = = the third term. 2 , R iR P = the rate at which electrical energy is dissipated in the resistance. , L L d UP dt = the rate at which the amount of energy stored in the inductor is changing. , CC d dt = the rate at which the amount of energy stored in the capacitor is changing. EVALUATE: The equation says that 0; RLC PPP + the net rate of change of energy in the circuit is zero. Note that at any given time one of or CL PP 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. IDENTIFY: The energy stored in a capacitor is 2 1 2 C UC v = . The energy stored in an inductor is 2 1 2 L i = . Energy conservation requires that the total stored energy be constant. SET UP: The current is a maximum when the charge on the capacitor is zero and the energy stored in the capacitor is zero. EXECUTE: (a) Initially 16.0 V v = and 0 i = . 0 L U = and 26 2 4 (5.00 10 F)(16.0 V) 6.40 10 J C v × = × . The total energy stored is 0.640 mJ . (b) The current is maximum when 0 q = and 0 C U = . 4 UU += × so 4 L U . 24 1 max 2 Li and 4 max 3 2(6.40 10 J) 0.584 A 3.75 10 H i × × . EVALUATE: The maximum charge on the capacitor is 80.0 C QC V μ = = . 30.57. IDENTIFY and SET UP: Use 2 1 2 V = (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. EXECUTE: 2 1 2 12.0 V; so 2 / 2(0.0160 J)/(12.0 V) 222 F C C C VU C V C U V = = 2 so (2 ) 2 fL f C LC 3500 Hz gives 9.31 H EVALUATE: f is in Hz and is in rad/s; we must be careful not to confuse the two. 30.58. IDENTIFY: Apply energy conservation to the circuit. SET UP: For a capacitor / VqC = and 2 Uq C = . For an inductor 2 1 2 i = EXECUTE: (a) 6 max 4 6.00 10 C 0.0240 V. 2.50 10 F Q V C × = × (b) 2 2 max 1 Q Li C = , so 6 3 max 4 1.55 10 A (0.0600 H)(2.50 10 F) Q i LC × = × × (c) 23 2 8 max max (0.0600 H)(1.55 10 A) 7.21 10 J. i × = ×

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30-16 Chapter 30 (d) If max 1 2 ii = then 8 max 1 1.80 10 J 4 L UU == × and ( ) 2 2 max (3/ 4) 3 42 2 C Q q CC = . This gives 6 3 5.20 10 C.
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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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891_PartUniversity Physics Solution - Inductance (b) i =...

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