Physics 4B. A positive value is obtained, so the current is counterclockwise around the circuit.

Transcription

1 Physcs 4B Solutons to Chapter 7 HW Chapter 7: Questons:, 8, 0 Problems:,,, 45, 48,,, 7, 9 Queston 7- (a) no (b) yes (c) all te Queston μc Queston 7-0, c;, a;, d; 4, b Problem 7- (a) Let be the current n the crcut and take t to be postve f t s to the left n R. We use Krchhoff s loop rule: R R = 0. We solve for : = R + R V. 0 V = = 050. A. 4.0Ω+ 80. Ω A postve value s obtaned, so the current s counterclockwse around the crcut. If s the current n a resstor R, then the power dsspated by that resstor s gven by P= R. (b) For R, P = R= (0.50 A) (4.0 Ω) =.0 W, (c) and for R, P = R = (0.50 A) (8.0 Ω) =.0 W. If s the current n a battery wth emf, then the battery supples energy at the rate P = provded the current and emf are n the same drecton. The battery absorbs energy at the rate P = f the current and emf are n opposte drectons. (d) For, P = = (0.50 A)( V) =.0 W (e) and for, P = = (0.50 A)(.0 V) =.0 W. (f) In battery the current s n the same drecton as the emf. Therefore, ths battery supples energy to the crcut; the battery s dschargng.

2 (g) The current n battery s opposte the drecton of the emf, so ths battery absorbs energy from the crcut. It s chargng. Problem 7- Let be the current n R and take t to be postve f t s to the rght. Let be the current n R and take t to be postve f t s upward. (a) When the loop rule s appled to the lower loop, the result s The equaton yelds R = V = = = A. R 00Ω (b) When t s appled to the upper loop, the result s The equaton gves R =. 0.0 V 5.0 V 4.0 V = = = R 50Ω 0.00 A, or = 0.00 A. The negatve sgn ndcates that the current n R s actually downward. (c) If V b s the potental at pont b, then the potental at pont a s V a = V b + +, so V a V b = + = 4.0 V V = 9.0 V. Problem 7- Frst, we note n V 4, that the voltage across R 4 s equal to the sum of the voltages across R 5 and R : V 4 = (R 5 +R )= (.40 A)(8.00 Ω Ω) =.8 V. The current through R 4 s then equal to 4 = V 4 /R 4 =.8 V/(.0 Ω) =.05 A. By the juncton rule, the current n R s so ts voltage s V = (.00 Ω)(.45 A) = 4.90 V. = 4 + =.05 A +.40 A =.45 A, The loop rule tells us the voltage across R s V = V + V 4 =.7 V (mplyng that the current through t s = V /(.00 Ω) = 0.85 A).

3 The juncton rule now gves the current n R as = + =.45 A A =. A, mplyng that the voltage across t s V = (. A)(.00 Ω) =. V. Therefore, by the loop rule, = V + V =. V +.7 V = 48. V. Problem 7-45 (a) We note that the R resstors occur n seres pars, contrbutng net resstance R n each branch where they appear. Snce = and R = R, from symmetry we know that the currents through and are the same: = =. Therefore, the current through s =. Then from V b V a = R = + (R )() we get 4.0V.0V 4R + R 4.0Ω +.0Ω = = = ( ) Therefore, the current through s = = 0.7 A. (b) The drecton of s downward. (c) The current through s = 0. A. (d) The drecton of s upward. (e) From part (a), we have = = 0. A. (f) The drecton of s also upward. (g) V a V b = R + = (0. A)(.0 Ω) V =. V. 0.A. Problem 7-48 (a) We use P = /R eq, where R eq (.0 Ω)( 4.00 Ω) R (.0 Ω)( 4.0 Ω ) + (.0 Ω ) R + ( 4.00 Ω) = 7.00 Ω+. R Put P = 0.0 W and = 4.0 V and solve for R: R = 9.5 Ω. (b) Snce P R eq, we must mnmze R eq, whch means R = 0. (c) Now we must maxmze R eq, or set R =. (d) Snce R eq, mn = 7.00 Ω, P max = /R eq, mn = (4.0 V) /7.00 Ω = 8. W. (e) Snce R eq, max = 7.00 Ω + (.0 Ω)(4.00 Ω)/(.0 Ω Ω) = 0.0 Ω, P mn = /R eq, max = (4.0 V) /0.0 Ω = 57. W.

4 Problem 7- (a) The voltage dfference V across the capactor s V(t) = ( e t/rc ). At t =.0 μs we have V(t) = 5.00 V, so 5.00 V = (.0 V)( e.0 μs/rc ), whch gves τ = (.0 μ s)/ln(/7) =.4 μs. (b) The capactance s C = τ/r = (.4 μs)/(5.0 kω) = pf. Problem 7- At t = 0 the capactor s completely uncharged and the current n the capactor branch s as t would be f the capactor were replaced by a wre. Let be the current n R and take t to be postve f t s to the rght. Let be the current n R and take t to be postve f t s downward. Let be the current n R and take t to be postve f t s downward. The juncton rule produces = +, the loop rule appled to the left-hand loop produces R R = 0, and the loop rule appled to the rght-hand loop produces R R =. 0 Snce the resstances are all the same we can smplfy the mathematcs by replacng R, R, and R wth R. (a) Solvng the three smultaneous equatons, we fnd c c h h. 0 V = = R Ω =. 0 A, (b) (c). 0 V = = = R 0.70 ( Ω) = = A A, and At t = the capactor s fully charged and the current n the capactor branch s 0. Thus, =, and the loop rule yelds R R = 0. (d) The soluton s. 0 V = = = R 0.70 ( Ω) 8. 0 A. (e) = = 8. 0 A.

5 (f) As stated before, the current n the capactor branch s = 0. We take the upper plate of the capactor to be postve. Ths s consstent wth current flowng nto that plate. The juncton equaton s = +, and the loop equatons are R R = 0 q R + R = 0. C We use the frst equaton to substtute for n the second and obtan R R = 0. Thus = ( R)/R. We substtute ths expresson nto the thrd equaton above to obtan (q/c) ( R) + (/) ( R/) = 0. Now we replace wth dq/dt to obtan Rdq q + =. dt C Ths s just lke the equaton for an RC seres crcut, except that the tme constant s τ = RC/ and the mpressed potental dfference s /. The soluton s The current n the capactor branch s The current n the center branch s c h RC. C q = e t dq dt R () t RC. = = e t and the potental dfference across R s t e e R R R R ( ) () t RC t RC = = = (g) For (h) For t t t RC V() t R = = ( e ). t RC 0, e = = and V ( ) = =. 0 V = V. t RC, e = 0 and V ( ) = =. 0 V =.0 0 V. () A plot of V as a functon of tme s shown n the followng graph.

6 Problem 7-7 (a) The four resstors R, R, R, and R 4 on the left reduce to R R R RR RR 4 eq = + 4 = + = 7.0 Ω+.0 Ω= 0 Ω R+ R R+ R4 Wth = 0 V across R eq the current there s =.0 A. (b) The three resstors on the rght reduce to RR (.0 Ω)(.0 Ω) R = R + R = + R = +.5 Ω=.0 Ω. +.0 Ω+.0 Ω 5 eq R5 R. Wth = 0 V across R eq the current there s 4 = 0 A. (c) By the juncton rule, = + 4 = A. (d) By symmetry, = =.5 A. (e) By the loop rule (proceedng clockwse), readly yelds 5 = 7.5 A. 0V 4 (.5 Ω) 5 (.0 Ω) = 0 Problem 7-9 The equvalent resstance of the seres par of R = R 4 =.0 Ω s R 4 = 4.0 Ω, and the equvalent resstance of the parallel par of R = R = 4.0 Ω s R =.0 Ω. Snce the voltage across R 4 must equal that across R : V = V R = R =

7 Ths relaton, plus the juncton rule condton I = + 4 =.00 A, leads to the soluton /.00 A = 4.0 A. It s clear by symmetry that = =.