Chapter 1 Electric Circuit Variables

Transcription

1 Chaper 1 Elecric Circui Variables Exercises Exercise Find he charge ha has enered an elemen by ime when i = A, 0. Assume q() = 0 for < Answer: q () = 2 C 3 2 i() = 8 4 A q () = idτ + q(0) = (8τ 4 τ) dτ 0 τ 2τ 2 C 0 + = = Exercise The oal charge ha has enered a circui elemen is q() = 4 sin 3 C when 0 and q() = 0 when < 0. Deermine he curren in his circui elemen for > 0. Answer: i () = d 4sin3= 12cos3 A d dq d i() = = 4sin3 = 12cos3 A d d Exercise Which of he hree currens, i 1 = 45 μa, i 2 = 0.03 ma, and i 3 = 25 4 A, is larges? Answer: i 3 is larges. i 1 = 45 μa = 45-6 A < i 2 = 0.03 ma =.03-3 A = 3-5 A < i 3 = 25-4 A 1-1

2 Exercise Figure E shows four circui elemens idenified by he leers A, B, C, and D. (a) Which of he devices supply 12 W? (b) Which of he devices absorb 12 W? (c) Wha is he value of he power received by device B? (d) Wha is he value of he power delivered by device B? (e) Wha is he value of he power delivered by device D? Answers: (a) B and C, (b) A and D, (c) 12 W, (c) 12 W, (e) 12 W + 4 V 2 V V 3 V + 6 A 2 A 4 A 3 A (A) (B) (C) (D) Figure E (a) B and C. The elemen volage and curren do no adhere o he passive convenion in B and C so he produc of he elemen volage and curren is he power supplied by hese elemens. (b) A and D. The elemen volage and curren adhere o he passive convenion in A and D so he produc of he elemen volage and curren is he power delivered o, or absorbed by hese elemens. (c) 12 W. The elemen volage and curren do no adhere o he passive convenion in B, so he produc of he elemen volage and curren is he power received by his elemen: (2 V)(6 A) = 12 W. The power supplied by he elemen is he negaive of he power delivered o he elemen, 12 W. (d) 12 W (e) 12 W. The elemen volage and curren adhere o he passive convenion in D, so he produc of he elemen volage and curren is he power received by his elemen: (3 V)(4 A) = 12 W. The power supplied by he elemen is he negaive of he power received o he elemen, 12 W. 1-2

3 Problems Secion 1-2 Elecric Circuis and Curren Flow P1.2.1 d ( ) ( ) i = e = 1.5e d 5 5 P ( ) ( ) ( ) ( τ ) τ τ τ τ τ 6 τ q = i d + q = e d + = d e d = + e C P q( ) = i( τ) dτ + q( 0) = 5sin 6 d 0 cos3 cos 6 C 0 τ τ + = τ = P () ( ) 0 0 C for 2 so q(2) = 0. q = iτ dτ = dτ = () ( τ) τ ( 2) 2 τ 2τ C for 2 4. In paricular, q(4) = 4 C. q = i d + q = d = = 2 2 () ( τ) τ ( ) τ τ 4 8 C for 4 8. In paricular, q(8) = 0 C. q = i d + q4 = 1d + 4= + 4= 4 4 () ( ) ( ) q = iτ dτ + q8 = 0dτ + 0= C for

4 P < 0 dq( ) i ( ) = i() = 3 0< < 2 d 2( 2) 2 e > 2 P C i = 460 A = 460 s C s mg = = = s min C 5 Silver deposied min mg g 1-4

5 P1.2-7 and 1 when 0< 4 i () = 1 when 2 q () = i( τ) dτ + q(0) = i( τ) dτ 0 0 since q(0) =0. When 0 < 4, we have When 4, we have q= 1 dτ = C q = i ( τ) dτ = 1 dτ + ( τ 1) dτ τ = τ τ = The skech of q() is shown o he righ:. C 1-5

6 Secion 1-3 Sysems of Unis P ( )( ) Δ q = i Δ = s = = A C 3.5 nc P Δ q 45 i = = = 9 3 Δ 5 6 = 9 μa P Soluion elecron 19 C 9 elecron 19 C i = 20 billion = s elecron s elecron 19 elecron C = s elecron 9 C = = na s P1.3-4 Soluion ma d = = i ( ) = q( ) = he slope of he q versus plo = 2 9 d = 6.7 = 6.7 ma 3 1-6

7 P μA dτ when 0 < < 80 ms 0 q = i d = A d when 80 ms < < 140 ms 6 3 ( ) ( τ) τ ( )( ) ( ) 0 μ τ 80 ms ( )( ) ( )( ) 6 ( 500 ) when 0 < < 80 ms 6 6 = ( 40 ) + ( 650 ) when 80 ms < < dτ 1 C when 140 ms < 40 ms While 0 < < 80 ms q() increases linearly from 0 o 40 μc and while 80 < < 140 ms q() decreases linearly from 40 o 0 μc. Here s he skech: 140 ms P μs ( ) ( ) q = iτ dτ = "area under he curve beween 300 μs and 1250 μs" 300 μs q pc = + = + = 2 ( ) ( ) ( )( ) ( ) 1-7

8 Secion 1-5 Power and Energy P1.5-1 (a) A and D. The elemen volage and curren do no adhere o he passive convenion in Figures P1.5- A and D so he produc of he elemen volage and curren is he power supplied by hese elemens. (b) B and C. The elemen volage and curren adhere o he passive convenion in Figures P1.5-1 B and C so he produc of he elemen volage and curren is he power delivered o, or absorbed by hese elemens. (c) 60 mw. The elemen volage and curren adhere o he passive convenion in Figure P1.5-1B, so he produc of he elemen volage and curren is he power received by his elemen: ( v)(6 ma) = 60 mw. The power supplied by he elemen is he negaive of he power received o he elemen, 60 mw. (d) 60 mw (e) 60 mw. The elemen volage and curren adhere o he passive convenion in Figure P1.5-1C, so he produc of he elemen volage and curren is he power received by his elemen: (12 V)(5 ma) = 60 mw. The power supplied by he elemen is he negaive of he power received o he elemen, 60 mw. P a.) q = id iδ ( )( )( ) b.) P= vi = ( 1 V)( A) = 10 W c.) = = A 2 hrs 3600s/hr = 7.2 C 0.12$ Cos = 1.1kW 2 hrs = $ kwhr 1-8

9 P ( )( ) P = 8V ma = 0.08 W Δw 220 W s Δ = = = P 0.08 W s P for 0 s: v = 30 V and i = = 2 A P = 30(2 ) = 60 W for 15 s: v() = + b v( ) = 30 V b= 80 V 5 v ( ) = and i ( ) = 2 A P= = W ( )( ) ( )( ) 2 30 for s: v= 5 V and i( ) = + b A i(25) = 0 b = 75 i( ) = A P = = W ( ) ( ) 0 15 Energy = P d = 60 d d d = = J

10 P a.) Assuming no more energy is delivered o he baery afer 5 hours (baery is fully charged). 6(3600) 6( τ ) w = Pd = vidτ = 2 11 dτ = 22+ τ = 661 J = 661 kj b.) 1 hr 15 Cos = 661kJ = s kwhr 1-

11 P p() v()() i ( 4cos3) = = sin 3 = ( sin 0 + sin 6) = sin W 1 p ( 0.5) = sin 3 = W 1 p () 1 = sin 6 = W Here is a MATLAB program o plo p(): clear 0=0; % iniial ime f=2; % final ime d=0.02; % ime incremen =0:d:f; % ime v=4*cos(3*); i=(1/12)*sin(3*); for k=1:lengh() p(k)=v(k)*i(k); end % device volage % device curren % power plo(,p) xlabel('ime, s'); ylabel('power, W') P The power is P=V I= = 0.5 W Nex, he energy is w = P = = 31.5J. 1-11

12 Secion 1.7 How Can We Check? P Noice ha he elemen volage and curren of each branch adhere o he passive convenion. The sum of he powers absorbed by each branch is: (-2V)(5A)+(6V)(2A)+(3V)(4A)+(4V)(-5A)+(1V)(6A) = -W + 12W + 12W -20W + 6W = 0 W The elemen volages and currens saisfy conservaion of energy and may be correc. P Noice ha he elemen volage and curren of some branches do no adhere o he passive convenion. The sum of he powers absorbed by each branch is: -(3V)(6A)+(3V)(2A)+ (5V)(2A)+(4V)(6A)+(-3V)(-3A)+(4V)(-3A) = -18W + 6W + W + 24W + 9W -12W 0 The elemen volages and currens do no saisfy conservaion of energy and canno be correc. 1-12

13 Design Problems DP 1-1 The volage may be as large as 20(1.25) = 25 V and he curren may be as large as (0.008)(1.25) = 0.01 A. The elemen needs o be able o absorb (25 V)(0.01 A) = 0.25 W coninuously. A Grade B elemen is adequae, bu wihou margin for error. Specify a Grade B device if you rus he esimaes of he maximum volage and curren and a Grade A device oherwise. DP 1-2 () = ( ) = ( ) p e e e e W Here is a MATLAB program o plo p(): Here is he plo: clear 0=0; ime f=1; ime d=0.02; incremen =0:d:f; % iniial % final % ime % ime v=20*(1-exp(-8*)); % device volage i=.030*exp(-8*); % device curren for k=1:lengh() p(k)=v(k)*i(k); end % power plo(,p) xlabel('ime, s'); ylabel('power, W') The circui elemen mus be able o absorb 0.15 W. 1-13