Homework+Ch+26 - [Fl-3" m 8354 Na'h mid at Wm Br" 8...

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Unformatted text preview: [Fl-3"] m 8354 Na'h mid at Wm Br" 8!: 5%] - a _ y: W ,_ P" I R ‘ R 1" R913 V: dim/min 1 V W we. 5': er v.3 m2 W. é Jfl I‘OJ‘I'IIUI3 w" a E: Gin-:5; M I Rafi? éfié: +83 R ‘1 11:3 _v = i =9 Real-3" I; Pfiz- M V ' -- *4 3: R -R .) - H - 3 ‘41. V [eyes a . a a ‘ z I. L Wm “WM W.e£&§—)l_1.=_\g (334 ‘- l Ref R\ R7. R, [9 *R3J‘V‘ I it -& 2 2: W" a __'f___,._. __ ___ z- I R4” R1. Ra Rt 3‘ m ER; I W," 1 i: W e) V‘é (K-V)%‘f-§W 3'2 vzrlwoav‘ g) ' *1! g) VMax= 56-6 V Hi circuii‘ Jam? fa, Ra MI 'HL I'l'él'l is 0P5", r a!) w t I“ 9...; :‘n V R; "D 39911,qu C4058 Rifles, N ’1 $ Vt (Until? «(rag (flu/1h an”. if 9 5.1.. 3"” “P filth} 13*2} 31:; I 5?: Ear-rt“! J 316. 1; \‘SB my J-MCiiOll Rule: I] “13+I3 22 an. .- 115). bop Rule: 11 V ‘27? '13! Laor (ecu) ‘99-- P [317- nvu [n.1,— QQBI, *DV— rL-iML‘Dll‘I. 2. o J 0 90 #0.} L... {c ) 6.0V I’ P w 6V- rIj—ME-l- an“ r “RV-[5:12]: In 2 "'/ 3 r" In W ware-6‘ It- Ia- I3 = o 134,-]; =7 DnIz-anIfisml-f W 35111, Mimi, = aw anyway 9, t - C Han—37.1111 = év )I2-fi+;%21’ => I9: a—g—(sA “11:3 3311, Hug;- (31! NH.) = aw L993 7, _3 ,5! a “24.1 asnl' 2'" 5'3” 175‘" J Irma—IL 26.34 Assume that the current flows in the direction labelled on the graph. If we solve the equations and find any current to be negative, it simply means that we assumed the wrong direction for the current and we will correct it. Look at the junction where the voltage V is at. Use the junction rule: I2=I1+Is Via—V 52—1/ Ezi R1 “ R3 Vi127V 94/ Ei 35 n 25 V:%V%T.86V 751 I1 : (51 — V)/R1 : 0.046A 12 : V/Rg : 0.163A 13 = (52 i V)/R3 = 0.118A Our current directions as labelled in the graph happen to be the correct direction for all three currents. (b) The internal resistances are in series with the batteries. If we look at the circuit, we see that the first battery’s internal resistor is in series with R1, the second battery’s is in series with R3. This means that with internal resistances, the equivalent circuit is one that looks exactly like the circuit we have, with the batteries being perfect, and with R1 being the original R1 plus r, and R3 similarly is the original R3 plus r. Therefore, we can simply use same equations as we did for part a, but with R1 = 25+1=26Q andR3=35+1=36Q I2:I1+Is Vigfiv 524/ ER R1 “ Rs V_127V 94/ @‘1233 W 26 7 V—szTBOilV 11 = (51 i V)/R1 = 0.046A 12 : V/Rg : 0.163A 13 : (52 — V)/R3 : 0.11m 26.64 I=V/R 5=IR+IT=V+%T Plug in the values for V and R: 9.7 5 Z W E?” 8.1 {S I W H?” Solve the system of equations, we get 5 = 112V and 7" = 5.39. 26.77 There are two possible answers because we don‘t know which direction the current is flowing in the circuit. First let‘s assume that the current given is flowing to the right. The voltage drop across the 4.0k resistor is V 2 IR 2 (3.10mA)(4.0kQ) = 12.4V, with the left side having higher potential since current is flowing towards the right. The current through the 8k resistor is thus I : V/R : DAV/81:9 : 1.55mA to the right (since left side has higher potential). Which means that the current through the 5k resistor is I : I1 + I2 : 3.10mA —|— 1.55mA : 4.65mA to the right. So the voltage drop across the 5k resistor is V 2 IR 2 (4.65mA)(5kQ) = 23.25V, with the left side being higher potential. The voltage drop across the 19 resistor is V 2 IR 2 (4.65mA)(1Q) = 0.00465V. Since the current flowing through it is towards the left, its right side has higher potential than the right side. Finally, putting everything together, we go around the circuit from a to b in counter— clockwise loop, and we get Va + 0.00465V + 121/ + 1241/ + 23.25v : V5 V5,, : Vb — Va : 47.65V Now let‘s consider the case that the 3.10mA is flowing through the 4k resistor towards the left. The voltage drop across the 4k resistor (and the 8k, consequently) are still the same 12.4V. But now, the right side has higher potential compared to the left side. The current through the 8k resistor is thus still 124/8000 = 1.55mA, but it’s towards the left. The current through the 5k resistor is still I : I1 + I2 : 4.65mA, but it‘s also towards the left. So voltage drop across the 5k resistor is still V : IR : (4.65mA)(5k§2) : 23.25V, but the right side has has potential now. Similarly, voltage drop across the 19 resistor is still V 2 IR 2 (4.65mA)(1Q) = 0.00465V, but as the current flowing through it is now towards the right, the left side has more potential now. So if we go around the same counterclockwise loop as before, from a to b, we get: Va 7 0.00465V + 12V 712.4V 7 23.25V = V}, In other words, except for the 12V battery, all other voltage drops have reversed their direction because the current reversed direction. And the answer in this case is: vim : vb — va : 70.00465 + 12 — 12.4 — 23.25 : 723.65V ...
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