Transcript ch2.5_SER
SERIES PARALLEL RESISTOR COMBINATIONS UP TO NOW WE HAVE STUDIED CIRCUITS THAT CAN BE ANALYZED WITH ONE APPLICATION OF KVL(SINGLE LOOP) OR KCL(SINGLE NODE-PAIR) WE HAVE ALSO SEEN THAT IN SOME SITUATIONS IT IS ADVANTAGEOUS TO COMBINE RESISTORS TO SIMPLIFY THE ANALYSIS OF A CIRCUIT NOW WE EXAMINE SOME MORE COMPLEX CIRCUITS WHERE WE CAN SIMPLIFY THE ANALYSIS USING THE TECHNIQUE OF COMBINING RESISTORS… … PLUS THE USE OF OHM’S LAW SERIES COMBINATIONS PARALLEL COMBINATION G p G1 G2 ... GN FIRST WE PRACTICE COMBINING RESISTORS 3k SERIES 6k||3k (10K,2K)SERIES 6k || 12k 4k 5k 3k 12k 3k || 6k 2k 12k 6k || (4k 2k ) 12k || 12k 6k If things get confusing… EXAMPLES COMBINATION SERIES-PARALLEL 9k If the drawing gets confusing… Redraw the reduced circuit and start again 18k || 9k 6k RESISTORS ARE IN SERIES IF THEY CARRY EXACTLY THE SAME CURRENT 6k 6k 10k RESISTORS ARE IN PARALLEL IF THEY ARE CONNECTED EXACTLY BETWEEN THE SAME TWO NODES AN “INVERSE SERIES PARALLEL COMBINATION” Given the final value Find a proper combination SIMPLE CASE VR MUST BE 600mV WHEN I 3A ONLY 0.1 RESISTORS ARE AVAILABLE REQUIRED R .6V 0.2 R 0.1 0.1 3A NOT SO SIMPLE CASE VR MUST BE 600mV WHEN I 9A ONLY 0.1 RESISTORS ARE AVAILABLE REQUIRED R .6V 0.0667 9A R EFFECT OF RESISTOR TOLERANCE NOMINAL RESISTOR VALUE : 2.7k RESISTOR TOLERANCE : 10% RANGES FOR CURRENT AND POWER? NOMINAL CURRENT : I 10 3.704 mA 2.7 NOMINAL POWER : P 10 2 2.7 _ 37.04 mW 10 3.367 mA MINIMUM POWER(VI min ) : 33.67 mW 1.1 2.7 10 4.115 mA MAXIMUM POWER : 41.15 mW 0.9 2.7 MINIMUM CURRENT : I min MAXIMUM CURRENT : I max THE RANGES FOR CURRENT AND POWER ARE DETERMINED BY THE TOLERANCE BUT THE PERCENTAGE OF CHANGE MAY BE DIFFERENT FROM THE PERCENTAGE OF TOLERANCE. THE RANGES MAY NOT EVEN BE SYMMETRIC CIRCUIT WITH SERIES-PARALLEL RESISTOR COMBINATIONS THE COMBINATION OF COMPONENTS CAN REDUCE THE COMPLEXITY OF A CIRCUIT AND RENDER IT SUITABLE FOR ANALYSIS USING THE BASIC TOOLS DEVELOPED SO FAR. COMBINING RESISTORS IN SERIES ELIMINATES ONE NODE FROM THE CIRCUIT. COMBINING RESISTORS IN PARALLEL ELIMINATES ONE LOOP FROM THE CIRCUIT GENERAL STRATEGY: •REDUCE COMPLEXITY UNTIL THE CIRCUIT BECOMES SIMPLE ENOUGH TO ANALYZE. •USE DATA FROM SIMPLIFIED CIRCUIT TO COMPUTE DESIRED VARIABLES IN ORIGINAL CIRCUIT - HENCE ONE MUST KEEP TRACK OF ANY RELATIONSHIP BETWEEN VARIABLES 4k || 12k 12k FIRST REDUCE IT TO A SINGLE LOOP CIRCUIT SECOND: “BACKTRACK” USING KVL, KCL OHM’S 6k I3 V OHM' S : I 2 a 6k KCL : I1 I 2 I 3 0 OHM' S : Vb 3k * I 3 …OTHER OPTIONS... 6k || 6k KCL : I 5 I 4 I 3 0 OHM' S : VC 3k * I 5 I1 12V 12k Va 3 (12) 39 12 I3 4 12 Vb 4k * I 4 I4 2k || 2k 1k VOLTAGE DIVIDER : VO LEARNING BY DOING 1k (3V ) 1V 1k 2k 1k 1k 2k CURRENT DIVIDER : I O 1k (3 A) 1A 1k 2k AN EXAMPLE OF “BACKTRACKING” 1.5mA I1 3mA V xz 6V 3V 1.5mA 1mA VO 36V 3V 0.5mA A STRATEGY. ALWAYS ASK: “WHAT ELSE CAN I COMPUTE?” Vb 6k * I 4 I3 Vb 3k I2 I3 I4 Va 2k * I 2 V xz Va Vb V I 5 xz 4k I1 I 2 I 5 VO 6k * I1 V xz 4k * I1 FIND VO 60k V1 6V FIND VS 2V 30k || 60k 20k STRATEGY : FIND V1 USE VOLTAGE DIVIDER 9V 12V VOLTAGE DIVIDER 20k VO V1 20k 40k I1 0.05mA 6V 120k VS 20k * 0.15mA 6V 20k 0.15mA 6V THIS IS AN INVERSE PROBLEM WHAT CAN BE COMPUTED? 20k + - V1 60k * 0.1mA V1 20k (12) 6V 20k 20k SERIES PARALLEL http://www.wiley.com/college/irwin/0470128690/animations/swf/D2Y.swf Y TRANSFORMATIONS THIS CIRCUIT HAS NO RESISTOR IN SERIES OR PARALLEL IF INSTEAD OF THIS WE COULD HAVE THIS THEN THE CIRCUIT WOULD BECOME LIKE THIS AND BE AMENABLE TO SERIES PARALLEL TRANSFORMATIONS Rab R2 || ( R1 R3 ) Y Rab Ra Rb Y Ra R1 Rb R1 R2 ( R1 R3 ) R R 1 2 R Ra Rb 3 Ra R R Ra b 3 R1 R2 R3 R1 R2 R3 Rb R2 RR R2 b 1 Rc R1 Rc REPLACE IN THE THIRD AND SOLVE FOR R1 R2 R3 R ( R R2 ) Rb Ra Rb Rb Rc Rc Ra Rb Rc 3 1 R1 R2 R3 R 1 R1 R2 R3 Rb R3 R1 Rc R R Rb Rc Rc Ra R1 R2 R3 R2 a b R1 ( R2 R3 ) Rc Rc Ra Y R1 R2 R3 R R Rb Rc Rc Ra R3 a b Ra SUBTRACT THE FIRST TWO THEN ADD TO THE THIRD TO GET Ra Y LEARNING EXAMPLE: APPLICATION OF WYE-DELTA TRANSFORMATION c R1 R3 R2 12k 6k 12k 6k 18k R1 R2 Ra R1 R2 R3 Rb R2 R3 R1 R2 R3 Rc R3 R1 R1 R2 R3 Y a b a c DELTA CONNECTION b COMPUTE IS REQ 6k 3k 9k || (2k 6k ) 10k IS 12V 1.2mA 12k ONE COULD ALSO USE A WYE - DELTA TRANSFORMATION ... LEARNING EXAMPLE CONVERT THIS Y INTO A DELTA? SHOULD KEEP THESE TWO NODES! IF WE CONVERT TO Y INTO A DELTA THERE ARE SERIES PARALLEL REDUCTIONS! R1 Ra Rb Rb Rc Rc Ra 3 *12k *12k 36k 12k Rb R2 Ra Rb Rb Rc Rc Ra Rc R3 Ra Rb Rb Rc Rc Ra Ra Y 36k 4mA 36k 36k THE RESULTING CIRCUIT IS A CURRENT DIVIDER 12k 12k V O CIRCUIT AFTER PARALLEL RESISTOR REDUCTION 36k ||12k 9k 4mA 36k IO 9k VO IO 36k 8 4mA mA 36k 18k 3 8 VO 9k I O 9k mA 24V 3 NOTICE THAT BY KEEPING THE FRACTION WE PRESERVE FULL NUMERICAL ACCURACY WYE DELTA