1. EEE 205 WS 2012 Part 4: Series & Parallel 118 R 1 R 2 R 3 Elements in series are joined at a common node at which no other elements are attached. The same current flows through the elements. Two resistors in series: R1R1 R2R2 Resistors not in series: RaRa RbRb RcRc R1R1 R2R2 RaRa RbRb RcRc R1R1 R2R2 4.1 Series Resistance & Parallel Resistance Series connection

2. EEE 205 WS 2012 Part 4: Series & Parallel 119 R1R1 R2R2 +–+– +v1–+v2–+v1–+v2– i Two Resistors in Series R1R1 R2R2 +–+– +v1–+v2–+v1–+v2– i KVL: v = v 1 + v 2 = R 1 i + R 2 i Same i! = (R 1 + R 2 ) i = R s i R 1 + R 2 +–+– i v v Memorize me! R s = R 1 + R 2 Very important formula!! i i v

3. EEE 205 WS 2012 Part 4: Series & Parallel 120 R1R1 R2R2 +–+– +v1–+v2–+vN–+v1–+v2–+vN– i N Resistors in Series KVL: v = v 1 + v 2 + … + v N = R 1 i + R 2 i + … + R N i = (R 1 + R 2 + … + R N ) i = R s i R s = R 1 + R 2 + … + R N Resistors in series “add.” v RNRN R 1 + R 2 + … + R N +–+– i v

4. EEE 205 WS 2012 Part 4: Series & Parallel 121 In the parallel connection, each of the resistors in parallel is connected to the same pair of nodes. The same voltage is across them. R1R1 R2R2 RbRb RdRd RaRa RcRc ReRe RfRf 2. None of these resistors are in parallel: Examples: 1. R 1 and R 2 are in parallel: node a node b R1R1 R2R2 Parallel Connection

5. EEE 205 WS 2012 Part 4: Series & Parallel 122 Two Resistors in Parallel KCL: i = i 1 + i 2 = v/R 1 + v/R 2 Same v! = (1/R 1 + 1/ R 2 ) v = 1/R eq v where 1/ R eq = 1/ R 1 + 1/ R 2 = (R 1 + R 2 ) / R 1 R 2  R eq = R 1 R 2 / (R 1 + R 2 ) Another very important formula!Also note that R//R  R/2 R eq +–+– v Memorize me! R1R1 R2R2 i1i1 i2i2 +v–+v– i i v +–+–

6. EEE 205 WS 2012 Part 4: Series & Parallel 123 Resistors in parallel do not add, but their corresponding conductances do: 1/ R 1 = G 1 1/ R 2 = G 2 1/ R eq = G eq R eq = R 1 R 2 / (R 1 + R 2 ) 1/ R eq = 1/ R 1 + 1/ R 2 G eq = G 1 + G 2 Compute the resistance of the parallel connection of R 1 = 6  & R 2 = 9 . The calculator-friendly way to make the computation is as follows: R 1 // R 2 = ( 6 –1 + 9 –1 ) –1 = (2.778) –1 = 3.6  Example Solution:

7. EEE 205 WS 2012 Part 4: Series & Parallel 124 How does R eq compare to the individual R’s connected in parallel? RaRa RbRb R eq R eq = R 1 R 2 / (R 1 + R 2 ) G eq = G 1 + G 2 G eq > G 1  1/ R eq > 1/R 1  R eq < R 1 G eq > G 2  1/ R eq > 1/R 2  R eq < R 2 Conclusion: The equivalent resistance R eq is smaller than either R 1 or R 2.

8. EEE 205 WS 2012 Part 4: Series & Parallel 125 100  25  For the resistor combination below verify that the equivalent resistor is smaller than either of the resistors connected in parallel. Compute the equivalent resistance in the calculator-friendly manner as follows: R eq = (100 –1 + 25 –1 ) –1 = 20  20  < 25  < 100  (as expected) Example Solution:

9. EEE 205 WS 2012 Part 4: Series & Parallel 126 N resistors in parallel Memorize me! R1R1 R2R2 i i1i1 i2i2 iNiN … … +v–+v– KCL: i = i 1 + i 2 + … + i N = G 1 v + G 2 v + … + G N v Same v! = (G 1 + G 2 + … + G N ) v = G eq v where G eq = G 1 + G 2 + … + G N Conductances in parallel “add.” 1/R eq = 1/R 1 + 1/R 2 + … + 1/R N or, in the “calculator-friendly” form: R eq = ( R 1 –1 + R 2 –1 + … + R N –1 ) –1 RNRN v +–+–

10. EEE 205 WS 2012 Part 4: Series & Parallel 127 Find the equivalent resistance looking in to the right of a-b. R eq 6 6 8 2 Solution: R eq 6 6 8 2 6 series 2 = 8 R eq 6 8 8 8 // 8 = 4 R eq 6 4 6 series 4 = 10 = R eq a b a b a b a b Example 1.

11. EEE 205 WS 2012 Part 4: Series & Parallel 128 Find R eq looking in from a-b with c-d open and with c-d shorted, and looking in from c-d with a-b open and with a-b shorted. 1080  360  720  abab c d Note that there are four separate problems here. 1. Find R eq with c-d open and looking in at a-b: 1080  360  720  1800  720 series 1080 = 1800 360 series 1080 = 1440 1800 // 1440 = 800 = R eq abab abab 1440  Example 2. Solution:

12. EEE 205 WS 2012 Part 4: Series & Parallel 129 2. Find R eq with c-d shorted and looking in at a-b: 1080  360  720  c d 540  240  720 // 360 = 240 1080 // 1080 = 540 240 series 540 = 780 = R eq abab abab 1080  360  720  c d 3. Find R eq with a-b open and looking in at c-d: 720 series 360 = 1080 1080 series 1080 1080 = 2160 2160  1080  1080 // 2160 = 720 = R eq Solution (cont.): c d

13. EEE 205 WS 2012 Part 4: Series & Parallel 130 1080  360  720  abab c d 4. Find R eq with a-b shorted and looking in at c-d: 720 // 1080 = 432 = R eq 1080  360  432  abab c d 360 // 1080 = 270 = R eq 270  432  c d 432 series 270 = 702  = R eq abab Solution (cont.):

14. EEE 205 WS 2012 Part 4: Series & Parallel 131 R1R1 R2R2 +–+– +v1–+v2–+v1–+v2– i v i = v / (R 1 + R 2 ) v 1 = R 1 i = R 1 v/(R 1 + R 2 ) = [R 1 /(R 1 + R 2 )] v v 2 = R 2 i = [R 2 /(R 1 + R 2 )] v The equations for v 1 and v 2 are “voltage divider” equations. The voltage v “divides” between R 1 and R 2 in direct proportion to the sizes of R 1 and R 2. 4.2 Voltage Division & Current Division Voltage Division (2 Resistors)

15. EEE 205 WS 2012 Part 4: Series & Parallel 132 Voltage Division (N Resistors) i = v / (R 1 + R 2 + … + R N ) k th resistor: v k = R k I, so that R1R1 R2R2 +–+– +v1–+v2–+vN–+v1–+v2–+vN– i v RNRN RkRk +vk–+vk– i Memorize me!

16. EEE 205 WS 2012 Part 4: Series & Parallel 133 Find the voltage across each resistor using the indicated polarities. 24 V +–+– 5  3  v 3 = [ 3 / ( 3 + 5 + 4 ) ] * 24 = 3 / 12 * 24 = 6 V v 5 = – 5 / 12 * 24 = – 10 V v 4 = 4 / 12 * 24 = 8 V Note the negative sign in v 5 ! As a check, verify that KVL is valid: –24 + 6 – (–10) + 8 = 0 0 = 0 Ok! + v 3 – 4  – v 4 + – v 5 + Example 3. Solution:

17. EEE 205 WS 2012 Part 4: Series & Parallel 134 Current Division (2 resistors) i = (G 1 + G 2 ) v v = [ 1 / (G 1 + G 2 ) ] i i 1 = G 1 v = [ G 1 / (G 1 + G 2 ) ] i i 2 = G 2 v = [ G 2 / (G 1 + G 2 ) ] i The equations for i 1 and i 2 are “current divider” equations. The current i “divides” between the two resistances R 1 and R 2 in direct proportion to the sizes of the corresponding conductances G 1 and G 2. R1R1 R2R2 i i1i1 i2i2 +v–+v–

18. EEE 205 WS 2012 Part 4: Series & Parallel 135 Current Division (2 resistors) Formula In Terms of R’s, [Instead of G’s] R1R1 R2R2 i i1i1 i2i2 +v–+v– i 1 = G 1 / (G 1 +G 2 ) ] i = 1/R 1 / [ (1/R 1 + 1/R 2 ) ] = [ R 2 / (R 1 + R 2 ) ] i i 2 = G 2 / (G 1 +G 2 ) ] i = 1/R 2 / [ (1/R 1 + 1/R 2 ) ] i = [ R 1 / (R 1 + R 2 ) ] i If R 1 is large (relative to R 2 ), then i 1 is small. If R 2 is large (relative to R 1 ), then i 2 is small. The larger current flows through the smaller resistor.

19. EEE 205 WS 2012 Part 4: Series & Parallel 136 Find i 1 and i 2 12  8 A i1i1 i2i2 4  i 1 = [ 4 / (12 + 4) ] x 8 = 2 A The larger resistor has the smaller current. i 2 = [ 12 / (12 + 4) ] x 8 = 6 A The smaller resistor has the larger current. Example 4. Solution:

20. EEE 205 WS 2012 Part 4: Series & Parallel 137 Current Division ( N resistors) i = (G 1 + G 2 + … + G N ) v v = [ 1 / (G 1 + G 2 + … + G N ) ] i For the k th resistor the current is: i k = G k v = [G k / (G 1 + G 2 + … + G N ) ] i The current divides in proportion to the conductances. The larger currents flow through the larger conductances (smaller resistances). R1R1 R2R2 i i1i1 i2i2 RNRN +v–+v– Same voltage! … … iNiN

21. EEE 205 WS 2012 Part 4: Series & Parallel 138 Find all the currents. 7.6  52.5 A 15.2  30.4  i1i1 i2i2 i3i3 i 1 = 7.6 –1 / ( 7.6 –1 + 15.2 –1 + 30.4 –1 ) x 52.5 = 30 A i 2 = 15.2 –1 / ( 7.6 –1 + 15.2 –1 + 30.4 –1 ) x 52.5 = 15 A i 3 = 30.4 –1 / (7.6 –1 + 15.2 –1 + 30.4 –1 ) x 52.5 = 7.5 A Note that on your calculator you can to use the (TI) x-1 key to enter the conductances, and the (TI) 2nd ENTRY key to repeat the formula for editing. Example 5. Solution:

22. EEE 205 WS 2012 Part 4: Series & Parallel 139 Find i. 100 V – + 7  – + 40 V 12  42  36  18  7 // 42 = 6  18 // 36 = 12 . The simplified circuit is: 100 V – + 6  40 V 12  Combining the sources and combining the resistors gives the following circuit: Applying KVL: – 60 + 30 i T = 0  i T = 2 A iTiT i 60 V – + 30  100 V 40 V – + +–+– Example 6. Solution:

23. EEE 205 WS 2012 Part 4: Series & Parallel 140 100 V – + 7  – + 40 V 12  42  36  18  i T =2 A i Using current division, i = [ 36 / (36 + 18) ] x 2 = 4/3 A [ Could also say i = 18 –1 / (18 –1 + 36 –1 ) x 2 ] 7  42  36  18  i 2 A Now figure out how much of the 2 A flows through the 18  : 2 A Example 6 Solution (cont.)

24. EEE 205 WS 2012 Part 4: Series & Parallel 141 Find i and i 1. 1  10  4  12 V 3  6  4  Solution: 1  10  4  + – 12 V 3  6  4  6  6 + 6 = 12  4 // 12 = 3  1  10  4  + – 12 V 3  4 + 3 + 3 = 10  10 // 10 = 5  i = 12 / 6 = 2 A i i1i1 i1i1 i i Example 7 + –

25. EEE 205 WS 2012 Part 4: Series & Parallel 142 1  10  + – 12 V 10  i = 2 A 1 A 1  10  4  + – 12 V 3  6  4  1 A i1i1 i = 2 A By current division: i 1 = – 4 –1 / (4 –1 + 12 –1 ) x 1 = – 0.75 A Example 7. Solution (cont.)

26. EEE 205 WS 2012 Part 4: Series & Parallel 143 Find R eq,a-b, i, and v. – 6  + – 36 V 9  30  72  36  +v–+v– i 10  First find R eq,a-b. Afterwards we can find i and v. R eq a b – 6  + – 36 V 9  30  72  36  +v–+v– i 10  R eq a b 6 // 30 // 0 = 0 72 // 9 = 8 Example 8. Solution:

27. EEE 205 WS 2012 Part 4: Series & Parallel 144 – + – 36 V 8  36  +v–+v– i 10  R eq a b R eq,a-b = 36 // 18 = 12 . i s = 36 / 12 = 3 A v = 8 / ( 8 + 10 ) x 36 = 16 V i = ( 18 –1 ) / ( 18 –1 + 36 –1 ) x 3 = 2 A Example 8. Solution (cont.)