SIMPLE ESISTIVE CICUIT ANALYSIS C.T. Pan 1 3.1 Series esistive Circuits 3. Parallel esistive Circuits 3.3 Voltage-divider and Current-divider Circuits 3.4 Measuring Current and Voltage 3.5 Measuring esistance-the Wheatstone Bridge 3.6 Delta-to to-wye (Pi-to to-tee) Equivalent Circuits C.T. Pan
3.1 Series esistive Circuits Linear resistors (Ohm s s law) = ρ l A ρ: : resistivity of material C.T. Pan 3 3.1 Series esistive Circuits example of fixed resistors wirewound type carbon film type A resistor with zero resistance is called short circuit. A resistor with infinite resistance is called open circuit. C.T. Pan 4
3.1 Series esistive Circuits Color code 黑 棕 紅 橙 黃 綠 藍 紫 灰 白 0 1 3 4 5 6 7 8 9 金銀, error 金 ± 5%, 銀 ± 10% 10-1 10 - no color ± 0% C.T. Pan 5 3.1 Series esistive Circuits A : yellow B : violet C : red D : gold 47 X 10 Ω ± 5% C.T. Pan 6
3.1 Series esistive Circuits red Yellow Green blue Violet ed Yellow 4 Green 5 blue 6 Violet 7 C.T. Pan 7 3.1 Series esistive Circuits a + v 1 - i 1 b v s i s + 1 v 3 - + v - i d i 3 3 c C.T. Pan 8
3.1 Series esistive Circuits High school solution = + + eq 1 3 vs eq C.T. Pan 9 3.1 Series esistive Circuits i = i = i = 1 3 v 1 = 1 i 1 v = i v 3 = 3 i 3 v s eq i s = -i 1 C.T. Pan 10
3.1 Series esistive Circuits v s a d i s i 3 + v - 1 1 + v - 3 3 i 1 + v - b c i N=4 nodes b=4 branches b method Problem: Find b unknowns i 1, i, i 3, i s C.T. Pan v 1, v, v 3, v s ( Passive sign convention ) 11 3.1 Series esistive Circuits Solution: Step1. KCL for N-1=3 N nodes node a i s = -i 1 node b i 1 = i node c i = i 3 node d i 3 = -i s (1) () (3) (4) Only N-1 KCL equations are independent! e.g. Eq (4) can be obtained from the previous three equations C.T. Pan 1
3.1 Series esistive Circuits Step. KVL equation v 1 + v + v 3 + (-vs)= 0 (5) Step3. Component models v 1 = 1 i 1 (6) v = i (7) v 3 = 3 i 3 (8) vs = given (9) There are b (=8) equations for b unknowns. Normally, there exists a unique solution. Unless the modeling is not accurate enough. C.T. Pan 13 3. Parallel esistive Circuits i s v s v 1 1 v 3 3 i1 i i 3 v High school solution 1 1 1 1 = + + eq 1 3 C.T. Pan 14
3. Parallel esistive Circuits vs i s eq vs is = = ( i1 + i + i3) i k = v s k eq, k=1,,3 Problem : Find b unknowns v 1, v, v 3, v s i 1, i, i 3, i s C.T. Pan 15 3. Parallel esistive Circuits Solution : Step1. KCL for N-1=1 1 node node a i s +i 1 +i +i 3 = 0 (1) Step. KVL equations loop A : v s +(-v 1 ) = 0 () loop B : v s +(-v ) = 0 (3) loop C : v s +(-v 3 ) = 0 (4) C.T. Pan 16
3. Parallel esistive Circuits Step3. Component models v 1 = 1 i 1 (5) v = i (6) v 3 = 3 i 3 (7) v s = given (8) There are b = 8 unknowns and there are 8 equations. One can find the unique solution. C.T. Pan 17 3.3 Voltage-divider and Current-divider Circuits The voltage-divider v v 1 = = 1 + 1 + 1 v v s s C.T. Pan 18
3.3 Voltage-divider and Current-divider Circuits The loading effect ( L ) L eq = + L eq vo = v + = 1 1 Ł eq s 1+ + L ł v s C.T. Pan 19 3.3 Voltage-divider and Current-divider Circuits Generalized case v j eq = n 1 j = i = k C.T. Pan 0 j eq v s
3.3 Voltage-divider and Current-divider Circuits The current divider i1 is 1 1 1 1 = + eq 1 i vk = i s eq k C.T. Pan 1 i i 1, = 1, v1 = = + 1 1 v 1 = = + 1 i i s s 3.3 Voltage-divider and Current-divider Circuits Generalized case i 1 s i i1 j i j n i n 1 v \ eq j = s n = i 1 k eq j 1 v j eq i j = = i C.T. Pan j s
3.4 Measuring Current and Voltage An ammeter is an instrument designed to measure current and must be placed in series with current being measured. Example : An analog ammeter based on the d Arsonval meter movement. C.T. Pan Picture from ELECTIC CICUITS 8th EDITION by Nilsson iedel 8th EDITION, 3 A dc ammeter equivalent circuit A The movable coil is characterized by a voltage rating and a current rating. e.g. a commercial meter movement is rated at 50mV and 1mA1 This means when the coil current is 1mA1 at full-scale position, the voltage drop across the coil is 50mV mv. C.T. Pan 4
A dc ammeter equivalent circuit An analog ammeter consists of a d Arsonval d movement in parallel with a resister A Example1 : A 50mV, 1mA1 d Arsonval movement is to be used in an ammeter of 1A1 rating. Determine A C.T. Pan 5 A dc ammeter equivalent circuit Solution : At full scale rating 1.0A 1mA = 50mV Loading effect due to ammeter or ( ) A 50 = 999 A Ω 50mV m = = 0.05 Ω 1A 50 50 999 m = = 0.05 Ω 50+50 999 C.T. Pan 6
Analog DC Ammeter To allow multiple ranges, shunt resistors are connected in parallel with the movement meter. + n I = I, current divider m fs A dc ammeter equivalent circuit n m I I -I m = n m fs m C.T. Pan 7 A dc ammeter equivalent circuit Example :Design an ammeter for the following multiple ranges (A) 0 ~ 1 A (B) 0 ~ 100 ma (C) 0 ~ 10 ma Assume that m =50Ω and I m = 1mA for the adopted d Arsonval movement meter. C.T. Pan 8
A dc ammeter equivalent circuit Ans : current divider I I = I = m fs n + I -I n m m n m fs m -3 10 50 (A) shunt resistance = ; 0.05 Ω 1 1A-1mA 1 50 50 (B) = = =0.505 Ω (100-1)mA 99 1 50 50 (C) = = =5.556Ω 3 10-1 9 C.T. Pan 9 A dc ammeter equivalent circuit An ideal ammeter has an equivalent resistance of 0 Ω and functions as a short circuit. C.T. Pan 30
A dc voltmeter equivalent circuit An analog voltmeter consists of a d Arsonvald movement in series with a series resister V. V The added V determines the full-scale reading of the voltmeter. C.T. Pan 31 A dc voltmeter equivalent circuit Example 3: A 50mV, 1mA1 d Arsonval movement is used in a voltmeter of 150V rating. Determine V. C.T. Pan 3
A dc voltmeter equivalent circuit Solution: From voltage divider formula 50 50mV= 150 V +50 v =149950 Ω v Loading effect of the meter m =149950+50=150 kω which is in parallel with the element to be measured. C.T. Pan 33 A dc voltmeter equivalent circuit V = I ( + ) fs fs n m Vfs = - n I fs m The design is based on the worst case which occurs when the full-scale current I fs =I m flows through the meters. C.T. Pan 34
A dc voltmeter equivalent circuit Example 4 : Design a voltmeter for the following multiple ranges (A) 0 ~ 1 V (B) 0 ~ 5 V (C) 0 ~ 100 V Assume the d Arsonval movement meter has m =kω with full scale I fs =100μA. C.T. Pan 35 A dc voltmeter equivalent circuit Ans : (A) V =I ( +) fs fs m 1 Vfs 1 = - = -000=8000Ω 1 m I 100μ fs 5 (B) = -000=48KΩ 100μ 100 (C) = -000=998KΩ 3 100μ C.T. Pan 36
3.5 Measuring esistance The Wheatstone Bridge The Wheatstone bridge circuit is used to measure resistance of mediumvalues :1Ω ~ 1MΩ, with an accuracy of about ± 0.1 % The Wheatstone bridge circuit C.T. Pan 37 3.5 Measuring esistance The Wheatstone Bridge current detector : a galvanometer X : the unknown resister C.T. Pan 38
3.5 Measuring esistance The Wheatstone Bridge Measuring X Adjust 3 until zero current in the galvanometer. C.T. Pan i = i, i = i 1 3 x V = V, V = V ab ac bd cd i=i, i =i 1 1 1 3 X = 1 3 X = X 3 1 39 3.5 Measuring esistance The Wheatstone Bridge 1 and : 1, 10, 100, 1000 Ω such that = 0.001~1000 1 in decimal steps 3 : adjustable in integral values of resistance from 1 to 11,000 Ω. Lower than 1Ω1 and higher than 1MΩ resistances are difficult to measure due to thermal heating effect and current leakage effect, respectively. C.T. Pan 40
3.6 Delta-to to-wye (Pi-to to-tee) Equivalent Circuits Delta (Δ)( ) or Pi (π)( ) circuit b a c Wye (Y) or Tee (T) circuit C.T. Pan 41 3.6 Delta-to to-wye (Pi-to to-tee) Equivalent Circuits Δ Y transformation c 1 b a 3 C.T. Pan 4
3.6 Delta-to to-wye (Pi-to to-tee) Equivalent Circuits = // ( + ) = + ab c a b 1 = // ( + ) = + bc a b c 3 = // ( + ) = + ca b c a 3 1 = + + b c 1 a b c c a = a + b + c a b = 3 a + b + c C.T. Pan 43 3.6 Delta-to to-wye (Pi-to to-tee) Equivalent Circuits Y to Δ transformation Similarly, given 1,, 3 ; one can find a, b, c = a = b = c + + 1 3 3 1 + + C.T. Pan 44 1 1 3 3 1 + + 1 3 3 1 3
3.6 Delta-to to-wye (Pi-to to-tee) Equivalent Circuits Special case, if 1 = = 3 =, then a = b = c =3 C.T. Pan 45 3.6 Delta-to to-wye (Pi-to to-tee) Equivalent Circuits Example : Use Y Δ transformation to find V in the following circuit. C.T. Pan 46
3.6 Delta-to to-wye (Pi-to to-tee) Equivalent Circuits Solution : C.T. Pan 47 3.6 Delta-to to-wye (Pi-to to-tee) Equivalent Circuits Solution : 0 10+ 10 5+ 5 0 35 = = Ω a 0 350 = = 35 Ω b 10 350 = = 70 Ω c 5 C.T. Pan 48
3.6 Delta-to to-wye (Pi-to to-tee) Equivalent Circuits 70 // 8 = 0 Ω + V - 35 Ω 35 // 105 = 15 Ω 35Ω V = A = 35V C.T. Pan 49 Summary Objective 1 : eview series and parallel resistive circuit solution approach. Objective : ecognize the b method by standard mathematical formulation. ( KCL + KVL + Ohm s s Law ) Objective 3 : Know how to use simple voltage-divider and current- divider concepts to solve simple circuit. C.T. Pan 50
Summary Objective 4 : Be able to determine the reading of ammeters and voltmeters. Objective 5 : Understand how to a Wheatstone bridge is used to measure resistance. Objective 6 : Know when and how to use delta-to to-wye equivalent circuits to solve a simple circuit. C.T. Pan 51 Assignment : Chapter Problems Problem : 3.14 3.4 3.30 3.33 3.50 3.64 Due within one week. C.T. Pan 5
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