RESISTIVE CIRCUITS SINGLE LOOP CIRCUIT ANALYSIS SINGLE LOOP

RESISTIVE CIRCUITS SINGLE LOOP CIRCUIT ANALYSIS SINGLE LOOP

RESISTIVE CIRCUITS SINGLE LOOP CIRCUIT ANALYSIS SINGLE LOOP CIRCUITS BACKGROUND: USING KVL AND KCL WE CAN WRITE ENOUGH EQUATIONS TO ANALYZE ANY LINEAR CIRCUIT. WE NOW START THE STUDY OF SYSTEMATIC, AND EFFICIENT, WAYS OF USING THE FUNDAMENTAL CIRCUIT LAWS a 1 2 b 6 b ra n ch e s 6 nodes 1 lo o p 3 c 4

WRITE 5 KCL EQS OR DETERMINE THE ONLY CURRENT FLOWING f 6 e 5 d ALL ELEM EN TS IN SER IES O N LY O N E CUR R EN T THE PLAN BEGIN WITH THE SIMPLEST ONE LOOP CIRCUIT EXTEND RESULTS TO MULTIPLE SOURCE AND MULTIPLE RESISTORS CIRCUITS IMPORTANT VOLTAGE DIVIDER EQUATIONS VOLTAGE DIVISION: THE SIMPLEST CASE KVL ON THIS LOOP

SUMMARY OF BASIC VOLTAGE DIVIDER v R1 R1 v (t ) R1 R2 EXAMPLE: VS 9V , R1 90k, R2 30k VOLUME CONTROL? R1 15k A PRACTICAL POWER APPLICATION HOW CAN ONE REDUCE THE LOSSES? THE CONCEPT OF EQUIVALENT CIRCUIT THE DIFFERENCE BETWEEN ELECTRIC CONNECTION AND PHYSICAL LAYOUT THIS CONCEPT WILL OFTEN BE USED TO SIMPLFY THE ANALYSIS OF CIRCUITS. WE INTRODUCE IT SOMETIMES, FOR PRACTICAL CONSTRUCTION

HERE WITH A VERY SIMPLE VOLTAGE DIVIDER REASONS, COMPONENTS THAT ARE ELECTRICALLY CONNECTED MAY BE PHYSICALLY QUITE APART i vS R1 i vS + - R2 i R1 R2 + - vS R1 R2

AS FAR AS THE CURRENT IS CONCERNED BOTH CIRCUITS ARE EQUIVALENT. THE ONE ON THE RIGHT HAS ONLY ONE RESISTOR SERIES COMBINATION OF RESISTORS R1 R2 R1 R2 IN ALL CASES THE RESISTORS ARE CONNECTED IN SERIES CONNECTOR SIDE ILLUSTRATING THE DIFFERENCE BETWEEN PHYSICAL LAYOUT AND ELECTRICAL CONNECTIONS PHYSICAL NODE PHYSICAL NODE SECTION OF 14.4 KB VOICE/DATA MODEM CORRESPONDING POINTS

COMPONENT SIDE FIRST GENERALIZATION: MULTIPLE SOURCES v2 v R1 Voltage sources in series can be algebraically added to form an equivalent source. v3 + - R1 + - + - v1

i(t) R2 + - v5 We select the reference direction to move along the path. vVoltage drops are subtracted from rise R2 + - KVL v4 R1

vR1 v2 v3 vR 2 v4 v5 v1 0 Collect all sources on one side v1 v2 v3 v4 v5 vR1 vR 2 v v eq R1 vR 2 veq + - R2 SECOND GENERALIZATION: MULTIPLE RESISTORS FIND I ,Vbd , P (30k ) APPLY KVL TO THIS LOOP APPLY KVL TO THIS LOOP

LOOP FORVbd Vbd 12 20[k ] I 0 (KVL) Vbd 10V POWER ON 30k RESISTOR P I 2 R ( 10 4 A) 2 (30 *103 ) 0.3mW v R Ri i i VOLTAGE DIVISION FOR MULTIPLE RESISTORS THE INVERSE VOLTAGE DIVIDER R1 VS + - R2 VO VOLTAGE DIVIDER

VO R2 VS R1 R2 "INVERSE" DIVIDER VS R1 R2 VO R2 COMPUTEVS " INVERSE" DIVIDER VS 220 20 458.3 500kV 220 Find I and Vbd APPLY KVL

TO THIS LOOP 6 80kI 12 40kI 0 I 0.05mA Vbd 40kI 12V 0 Vbd 10V If Vad = 3V, find VS 3V INVERSE DIVIDER PROBLEM 25 15 20 VS 3 9V 20

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