Series/Parallel Resistor Reduction

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SERIES PARALLEL RESISTOR COMBINATIONS. UP TO NOW WE HAVE STUDIED CIRCUITS THAT. CAN BE ANALYZED WITH ONE APPLICATION OF.
RESISTIVE CIRCUITS

• SERIES/PARALLEL RESISTOR COMBINATIONS - A TECHNIQUE

TO REDUCE THE COMPLEXITY OF SOME CIRCUITS • LEARN TO ANALYZE THE SIMPLEST CIRCUITS

• THE VOLTAGE DIVIDER • THE CURRENT DIVIDER • WYE - DELTA TRANSFORMATION - A TECHNIQUE TO REDUCE

COMMON RESISTOR CONNECTIONS THAT ARE NEITHER SERIES NOR PARALLEL

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 + ... + G N

FIRST WE PRACTICE COMBINING RESISTORS

3k SERIES 6k||3k (10K,2K)SERIES

6k || 12k = 4k

5kΩ 3k

12k

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

EFFECT OF RESISTOR TOLERANCE

NOMINAL RESISTOR VALUE : 2.7kΩ RESISTOR TOLERANCE : 10% RANGES FOR CURRENT AND POWER?



10 NOMINAL CURRENT : I = = 3.704 mA 2.7

(10)

2.7

_

= 37.04 mW

10 = 3.367 mA MINIMUM POWER(VImin ) : 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

NOMINAL POWER : P =

2

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) 3+9

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

Y − Δ 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 )

Δ →Y Rab = Ra + Rb

Y →Δ

Ra R1 Rb R1 R2 ( R1 + R3 ) R1 R2 = ⇒ = R Ra + Rb = 3 Ra = R R Ra 3 b 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 = Δ →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 −Δ

LEARNING EXAMPLE: APPLICATION OF WYE-DELTA TRANSFORMATION

COMPUTE IS

c R1

a

c

DELTA CONNECTION

=

R3

12k × 6k 12k + 6k + 18k

R2

R1 R2 Ra = R1 + R2 + R3 Rb =

R2 R3 R1 + R2 + R3

Rc =

R3 R1 R1 + R2 + R3

Δ →Y

a

REQ = 6k + (3k + 9k ) || (2k + 4k ) = 10k

IS =

12V = 1.2mA 10k

ONE COULD ALSO USE A WYE - DELTA TRANSFORMATION ...