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 ...
