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SINGLE LOOP CIRCUITS

KVL ON THIS LOOP. VOLTAGE DIVISION: THE SIMPLEST CASE. 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. WRITE 5 KCL EQS

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SINGLE LOOP CIRCUITS

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  1. KVL ON THIS LOOP VOLTAGE DIVISION: THE SIMPLEST CASE 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 WRITE 5 KCL EQS OR DETERMINE THE ONLY CURRENT FLOWING • THE PLAN • BEGIN WITH THE SIMPLEST ONE LOOP CIRCUIT • EXTEND RESULTS TO MULTIPLE SOURCE • AND MULTIPLE RESISTORS CIRCUITS IMPORTANT VOLTAGE DIVIDER EQUATIONS

  2. SUMMARY OF BASIC VOLTAGE DIVIDER A “PRACTICAL” POWER APPLICATION VOLUME CONTROL? HOW CAN ONE REDUCE THE LOSSES?

  3. 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 HERE WITH A VERY SIMPLE VOLTAGE DIVIDER SOMETIMES, FOR PRACTICAL CONSTRUCTION REASONS, COMPONENTS THAT ARE ELECTRICALLY CONNECTED MAY BE PHYSICALLY QUITE APART AS FAR AS THE CURRENT IS CONCERNED BOTH CIRCUITS ARE EQUIVALENT. THE ONE ON THE RIGHT HAS ONLY ONE RESISTOR IN ALL CASES THE RESISTORS ARE CONNECTED IN SERIES

  4. PHYSICAL NODE PHYSICAL NODE CONNECTOR SIDE ILLUSTRATING THE DIFFERENCE BETWEEN PHYSICAL LAYOUT AND ELECTRICAL CONNECTIONS SECTION OF 14.4 KB VOICE/DATA MODEM CORRESPONDING POINTS COMPONENT SIDE

  5. FIRST GENERALIZATION: MULTIPLE SOURCES Voltage sources in series can be algebraically added to form an equivalent source. We select the reference direction to move along the path. Voltage drops are subtracted from rises i(t) KVL Collect all sources on one side

  6. APPLY KVL TO THIS LOOP APPLY KVL TO THIS LOOP SECOND GENERALIZATION: MULTIPLE RESISTORS VOLTAGE DIVISION FOR MULTIPLE RESISTORS

  7. APPLY KVL TO THIS LOOP THE “INVERSE” VOLTAGE DIVIDER INVERSE DIVIDER PROBLEM

  8. Notice use of passive sign convention Knowing the current one can compute ALL the remaining voltages and powers

  9. EXAMPLE DETERMINE I USING KVL 12V

  10. KVL HERE OR KVL HERE APPLY KVL TO THIS LOOP EXAMPLE Sometimes you may want to vary a bit

  11. IN PRACTICE NODES MAY ASSUME STRANGE FORMS THIS ELEMENT IS INACTVE (SHORT-CIRCUITED) LOW DISTORTION POWER AMPLIFIER SINGLE NODE-PAIR CIRCUITS THESE CIRCUITS ARE CHARACTERIZED BY ALL THE ELMENTS HAVING THE SAME VOLTAGE ACROSS THEM - THEY ARE IN PARALLEL

  12. LOW VOLTAGE POWER SUPPLY FOR CRT - PARTIAL VIEW SAMPLE PHYSICAL NODES COMPONENT SIDE CONNECTION SIDE

  13. THE CURRENT DIVISION APPLY KCL USE OHM’S LAW TO REPLACE CURRENTS DEFINE “PARALLEL RESISTANCE COMBINATION” BASIC CURRENT DIVIDER THE CURRENT i(t) ENTERS THE NODE AND SPLITS - IT IS DIVIDED BETWEEN THE CURRENTS i1(t) AND i2(t)

  14. WHEN IN DOUBT… REDRAW THE CIRCUIT TO HIGHLIGHT ELECTRICAL CONNECTIONS!! IS EASIER TO SEE THE DIVIDER

  15. POWER PER SPEAKER LEARNING EXTENSION - CURRENT DIVIDER CAR STEREO AND CIRCUIT MODEL THERE IS MORE THAN ONE OPTION TO COMPUTE I2

  16. APPLY KCL TO THIS NODE DEFINE “PARALLEL RESISTANCE COMBINATION” FIRST GENERALIZATION: MULTIPLE SOURCES

  17. APPLY KCL TO THIS NODE Ohm’s Law at every resistor General current divider SECOND GENERALIZATION: MULTIPLE RESISTORS

  18. Notice use of passive sign convention Once v(t) is known all other variables can be determined; e.g.,

  19. ` 20k||5k General current divider

  20. FIND THE CURRENT NOTICE THE MINUS SIGN COMBINE RESISTORS COMBINE THE SOURCES STRATEGY: CONVERT THE PROBLEM INTO A BASIC CURRENT DIVIDER BY COMBINING SOURCES AND RESISTORS. THE NEXT SECTION EXPLORES IN MORE DETAIL THE IDEA OF COMBINING RESISTORS

  21. I2 I1 9mA I1 I1 I2 I2 DIFFERENT LOOKS FOR THE SAME ELECTRIC CIRCUIT

  22. I1 I2 I1 REDRAWING A CIRCUIT MAY, SOMETIMES, HELP TO VISUALIZE BETTER THE ELECTRICAL CONNECTIONS I2

  23. Determine power delivered by source + V _

  24. ONE EQUATION, TWO UNKNOWNS. CONTROLLING VARIABLE PROVIDES EXTRA EQUATION KVL REPLACE AND SOLVE FOR THE CURRENT UNITS ARE EXPLICIT USE OHM’S LAW CIRCUITS WITH DEPENDENT SOURCES GENERAL STRATEGY TREAT DEPENDENT SOURCES AS REGULAR SOURCES AND ADD ONE MORE EQUATION FOR THE CONTROLLING VARIABLE A CONVENTION ABOUT DEPENDENT SOURCES. UNLESS OTHERWISE SPECIFIED THE CURRENT AND VOLTAGE VARIABLES ARE ASSUMED IN SI UNITS OF Amps AND Volts FOR THIS EXAMPLE THE MULTIPLIER MUST HAVE UNITS OF OHM A PLAN: SINGLE LOOP CIRCUIT. USE KVL TO DETERMINE CURRENT

  25. KCL TO THIS NODE. THE DEPENDENT SOURCE IS JUST ANOTHER SOURCE THE EQUATION FOR THE CONTROLLING VARIABLE PROVIDES THE ADDITIONAL EQUATION ALGEBRAICALLY, THERE ARE TWO UNKNOWNS AND JUST ONE EQUATION SUBSTITUTION OF I_0 YIELDS VOLTAGE DIVIDER NOTICE THE CLEVER WAY OF WRITING mA TO HAVE VOLTS IN ALL NUMERATORS AND THE SAME UNITS IN DENOMINATOR A PLAN: IF V_s IS KNOWN V_0 CAN BE DETERMINED USING VOLTAGE DIVIDER. TO FIND V_s WE HAVE A SINGLE NODE-PAIR CIRCUIT

  26. THE DEPENDENT SOURCE IS ONE MORE VOLTAGE SOURCE KVL TO THIS LOOP THE EQUATION FOR THE CONTROLLING VARIABLE PROVIDES THE ADDITIONAL EQUATION REPLACE AND SOLVE FOR CURRENT I … AND FINALLY A PLAN: ONE LOOP PROBLEM. FIND THE CURRENT THEN USE OHM’S LAW.

  27. KCL KVL KVL KCL A PLAN: ONE LOOP ON THE LEFT - KVL ONE NODE-PAIR ON RIGHT - KCL ALSO A VOLTAGE DIVIDER

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