Lecture 2 - Simple Resistive Circuits, Lecture notes of Electronics

Download Lecture 2 - Simple Resistive Circuits and more Lecture notes Electronics in PDF only on Docsity! SEE Spring 2019 Mai Linh Lecture #2: Resistive Circuits Textbook: Electric Circuits James W. Nilsson & Susan A. Riedel 9th Edition. link: http://blackboard.hcmiu.edu.vn/ to download materials (Chapter 3) SEE Spring 2019 Mai Linh Practical Perspective Resistive Touch Screens 2 What is a resistive touchscreen? Resistive touch screens consist of a glass or acrylic panel that is coated with electrically conductive & resistive layers made with indium tin oxide (ITO). The thin layers are separated by invisible spacers. Examples: smartphones with resistive touchscreens: LG Optimus LG GW620 Sony Ericsson Vivaz Nokia N97 mini Nokia N900 SEE Spring 2019 Resistors in series (cont.) * In general, if & resistors are connected in series, the equivalent single resistor has a resistance equal to the sum of the « resistances, or Kk R,, = R= R,+R,+...4R, i=l Uy ZR => v, 2 h Ry Ry Rs h Mai Linh SEE Spring 2019 Resistors in parallel | I, =i +i) +i; ++ >>, Mai Linh iT -7j -_—j7 -—7 —+ ‘ UR, =i,8, =i, =i,Ry =v, When two elements connect at a single node pair, they are said to be connected in parallel. Parallel-connected circuit elements have the same voltage across their terminals. Applying KCL Ry SE Rs RRs * From Ohm’‘s law x / ‘ Same node “No elements connected Therefore \ between nodes” Vv. ; v . v . ry . =— 1, = 2,1,= — &i,= — 4% R, R, R, R, + V, Lt. Rt, vv Vv ©) : 7 + 5 SEE Spring 2019 Resistors in parallel (cont.) 1 41 /1 #1 \ = isalstoteet | * Ry HR (ROR R, | b Gy = YG, =(G,+G,+...4+G,) i=l *« Special Case (two resistors in parallel) 1 1 1 Re a +» R, RR, RR, 34 $2 Rig = Aik b p—__4__] R +R, N Mai Linh SEE Spring 2019 Assessing Objective 1 « Find (a) v, (b) power delivered to the circuit by the current source, ~ and (c) the power dissipated in the 10 0 resistor, Ans.: eThe 6Qisinseries with the 10 9, 3A Ry = 6+10 =16Q eThe 16 9 is in parallel with the 64 0, 1 1664 3A = =1282 16464 Mai Linh toed 100 1) : z00 eRe =12.8Q 10 SEF Snrina 2019 Example (cont.) «The 12.8 9 is in series with the 7.2, R,,3 = 7.2+12.8 = 202 «The 30 9 is in parallel | with the 20 9, sa(f) , g Rua _ 30x20 I ** 30+20 : e Now applying Ohm's Laws V=IR,,, = §*12° =60V =129 iy «Power delivered by the current source p=iv=5*60" =300W Mai Linh 11 SEF Snrina 2019 Example (cont.) po Vi9q = 347.2% =21.6V Vingg = 3° X12.89 =38.4V Mai Linh . 38.4" Issn 7 =0.6A , 38.4" igang = ie =2.4A 384 ¥ 640 ; 60" j ing =a =2A 3 | 30° 300 g200 so | boo = a0e =3A 7T2Q Ne a, lla, 0-64) | F244) 17 2a| 34 38.4V 1622 SEE Snrinan 2N19 Problems * Inthe circuit shown, find the equivalent resistance &,,, and the power delivered by the source. 450 . _ . et => = 00 hort Circuit) Regs = 10+ 20 = 300 P30 1 5x = =100 = =160 715430 : we ests Rags = 30+ 20 = 500 | Reg =24416+10=500 30x60 (gox10°} = = 200 : aoe LY? r 1 -30+60 Peony 50 128uW 15 Mai Linh SEE Spring 2019 The voltage-divider circuit Apply KVL => 4 vy, =iR, +iR, R, v, ix V, _ R+R Ys R . . R R, ae V3 vy, =iR, =v, ——— & v,=ik,=v,—— - R,+R, ° “ "RAR, Example: If the resistors used in the circuit have a tolerance of 210%. Find v,,,,. and v,,.., Ans.!- vov(! 1001.1 it v, (max) = 100 ————-—. = $3.02 V 100*1.14+25«0.9 1000.9 v, (min) = 100 —- — = 76.60 V 100«0.9+25*1.1 16 Mai Linh SEE Spring 2019 The current-divider circuit R,/R; ==> R= RR TORAR, Ohm's Law RR v=i,R, =1,R, =i,8,, =i, = - " " R+R, i= R, i, & = KR, i R+R, * RRS Escape Find the power dissipated in the 6 9 resistor ANS 67404160 ven Ry = 7p tl 6=40 wa) 16:12 ac Zen + Using Current Divider i,= 16 10=8A 16+4 1D Al { aad] i Using Current Divider ne ] . 4 2 i, =——8=32A Pi=i =6144W 6 446 6 sattsny Mai Linh 17 SEE Spring 2019 Voltage Division and Current Division ¥ ¥ = =— ] Ry Ry R,+R,+...+R, Reg 7 Vv; =iR, Circuit |e * Rez; i Vv; = R, Vv Ry Rat - i v=i(R, || R,|l...|| R,J=iR, * V=i,R; Circuit K, Ry Ry 4! Root Ryu I Res; 7 R I 20 Mai Linh SEE Spring 2019 Example 3.4 Use current division to find i, and voltage division to find v, Ans. :- Current Division i, = s =2A Ohm's Law Vy = 2494° =48V Voltage Division Qo Vy = 48" > =18V Mai Linh 0 3on 44.9) 10 f) 40.0 yi 4 ONE WHF v 4 30 0 v, 21 SEE Spring 2019 Assessing Objective 3 Use voltage division & current division to find (a) v,, (BD) isao & fsgar (C) Prop QO 500 Ans.:- 1 ' ' (a) R,, =———— =102 8 dthte et -60” © __a9y TG ° 40410470 . 60" (b) ‘400 = T3598 =0.5A i _i. Ren _ 9 5: 110 9 16674 70 400 Reon 30 (c) v, =60" p10 sy ioe yp SP 50 4.1667 V Rn" 40410470. = * ~ 50410 2 V, Pog = = 0.3472 W 300 22 Mai Linh SEE Spring 2019 Measuring Resistance-The Wheatstone Bridge « To find R,, the value of R,; is adjusted until there is no current in the meter. i, =0 Means balanced bridge. & V,,=0 “same potential at a as b” KCL 4=i, & i, =i, ; \ _ ss KVL ,R, =i,R, & LR, =1,R, ¥ 1R, =1,R, i RR, R, 4 BR on, lp Bp qy 1 3 R, 25 Mai Linh SEE Spring 2019 The A-to-Y transformation R. a b a — Rk, Ri, R, c « The resistance between terminals a and b must be the same whether we use A-connected set or the Y-connected circuit. = R(R,+R,) -RiR, i R = a 5 pn eR tRR + RR, R,+R, +R, - my at + c a R, = R,(Ry +R) _ p +R R= R.R, R, = RR, +R,R; + RR, R,+R, +R, " ! R, +R, +R, R, _ R,(R, + R,) —-R-=R. if R= RR, R= RAR, +R + RA, “OR, +R,+R 9) 7} 7 +R, +R, ° R, Mai Linh SEE Spring 2019 Example 3.7 Find the current and power supplied by the 40 V source. Ans.: We can convert A(100, 125, 25 9) or A(25,40,37.5 9) 100x125 100+125+25 _:125%25 21004125425 100x250 1004125425 — 9543 (10+40)*(12.5+37-5) _ 99.4 (10+ 40) + (12.5+37.5) ove Ly on j---=05A P=0.5* x80" =40 W L_ — Rg =5+50+ 27 Mai Linh SEE Spring 2019 Problem Specify the value of the resistors in the circuit to meet the following design criteria: i,=8 mA; v,=4 V, i,=2i,; 7,=10i,; and ITT Ans. :- t , . 1 1 1 1 1 Vv, =1 Ret tt gee RR, OR, Rs Ry : 1 . . 4” =8x10°( ) i vet | Rh R's als 3 dytiiit * T RR OR ORS 1,4 pop ticoxto? Pd | RR, RR, ' Vv Re. i =, =e; _ 7 Ra j , R, =2R, qj = = I 1 2 z R g RR; 1 2 i, = 10, + R, =10R, +44 + + == =2x107 i,=i, + R,=R, R, 2R, 20R, 20R, 20R, R, =8000 R,=16kO) |R, =R, =16kO Mai Linh SEE Spring 2019 Problem Find w,? Ans. ¢- Using current dividers - q 15 mA t 200 +1000 = Sim A 300 +3004 200+1000 y 7,=10mA . 300+ 300 - i, = lima ~ 3004+300+ 200+1000 i, =5mA y =10x10* x300=3V, Vy =5x10%x1000=5V Applying KVL v,+¥,—-v,) =O cout» v Mai Linh 31 SEE Spring 2019 Problem Find (a) v,, (b) power dissipated in 20 ©. Ans:- v,=144V 200 Pye = 28.8 W 200 32 Mai Linh