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R 1 R 2 R 3

R 1 R 2 R 3

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R 1 R 2 R 3

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  1. 4.1Series Resistance & Parallel Resistance Elements in series are joined at a common node at which no other elements are attached. The same current flows through the elements. Series connection R1R2R3 Two resistors in series: Resistors not in series: R1 Ra Rb R2 Rc Ra Rb R1 R2 Rc R1 R2 EEE 205 WS 2012 Part 4: Series & Parallel

  2. Two Resistors in Series + v1 – + v2 – + v1 – + v2 – i i R1 R1 v v i + – + – R2 R2 i • KVL: • v = v1 + v2 • = R1i + R2iSame i! • = (R1 + R2) i • = Rsi Memorize me! • Rs = R1 + R2 • Very important formula!! i R1 + R2 v + – EEE 205 WS 2012 Part 4: Series & Parallel

  3. N Resistors in Series + v1 – + v2 – + vN – i R1 i R2 R1 + R2 + … +RN v v + – + – RN • KVL: • v = v1 + v2 + … + vN • = R1i + R2i + … + RNi • = (R1 + R2 + … + RN) i • = Rs i • Rs = R1 + R2 + … + RN • Resistors in series “add.” EEE 205 WS 2012 Part 4: Series & Parallel

  4. In the parallel connection, each of the resistors in parallel is connected to the same pair of nodes. The same voltage is across them. Parallel Connection node a node b R2 R1 Examples: 1. R1 and R2 are in parallel: R1 R2 2. None of these resistors are in parallel: Ra Rb Rc Rf Rd Re EEE 205 WS 2012 Part 4: Series & Parallel

  5. Two Resistors in Parallel i i2 + v – v R1 R2 i1 + – • KCL: • i = i1 + i2 • = v/R1 + v/R2Same v! • = (1/R1 + 1/ R2) v • = 1/Req v • where • 1/ Req = 1/ R1 + 1/ R2 • = (R1 + R2) / R1R2 •  Req = R1R2 / (R1 + R2) • Another very important formula!Also note that R//R  R/2 i Memorize me! Req v + – Memorize me! EEE 205 WS 2012 Part 4: Series & Parallel

  6. Resistors in parallel do not add, but their corresponding conductances do: • 1/ R1 = G1 • 1/ R2 = G2 • 1/ Req = Geq • Req = R1R2 / (R1 + R2) • 1/ Req = 1/ R1 + 1/ R2 • Geq= G1 + G2 Example Compute the resistance of the parallel connection of R1 = 6 & R2 = 9 . The calculator-friendly way to make the computation is as follows: R1 // R2 = ( 6–1 + 9–1)–1 = (2.778)–1 = 3.6  Solution: EEE 205 WS 2012 Part 4: Series & Parallel

  7. How does Req compare to the individual R’s connected in parallel? Ra Req Rb • Req = R1R2/ (R1 + R2) • Geq = G1 + G2 • Geq > G1 1/ Req > 1/R1 Req < R1 • Geq > G2 1/ Req > 1/R2 Req < R2 • Conclusion: The equivalent resistance Req is smaller than eitherR1orR2. EEE 205 WS 2012 Part 4: Series & Parallel

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

  9. N resistors in parallel i … + v – v R1 R2 i1 i2 RN iN + – … • KCL: • i = i1 + i2+ … + iN • = G1v + G2v + … + GNvSame v! • = (G1 + G2 + … + GN) v • = Geq v • where • Geq = G1 + G2 + … + GN • Conductances in parallel “add.” • 1/Req = 1/R1 + 1/R2 + … + 1/RN • or, in the “calculator-friendly” form: • Req = ( R1–1 + R2–1 + … + RN–1)–1 Memorize me! EEE 205 WS 2012 Part 4: Series & Parallel

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

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

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

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

  14. 4.2 Voltage Division & Current Division Voltage Division (2 Resistors) i = v / (R1 + R2) v1 = R1i = R1 v/(R1+ R2) = [R1/(R1 + R2)] v v2 = R2i = [R2/(R1 + R2)] v + v1 – + v2 – i R1 v + – R2 The equations for v1and v2are “voltage divider” equations. The voltage v “divides” between R1and R2in direct proportion to the sizes of R1and R2. EEE 205 WS 2012 Part 4: Series & Parallel

  15. Voltage Division (N Resistors) + v1 – + v2 – + vN – i R1 + vk – i R2 v Rk + – RN i = v / (R1 + R2+ … + RN) kth resistor: vk = RkI, so that Memorize me! EEE 205 WS 2012 Part 4: Series & Parallel

  16. Find the voltage across each resistor using the indicated polarities. Example 3. 3  + v3 – – v5 + 5  + – 24 V – v4 + 4  Solution: • v3 = [ 3 / ( 3 + 5 + 4 ) ] * 24 • = 3 / 12 * 24 = 6 V • v5 = –5 / 12 * 24 = – 10 V • v4 = 4 / 12 * 24 = 8 V • Note the negative sign in v5! • As a check, verify that KVL is valid: • –24 + 6 – (–10) + 8 = 0 • 0 = 0 Ok! EEE 205 WS 2012 Part 4: Series & Parallel

  17. Current Division (2 resistors) i2 + v – R1 R2 i1 i • i = (G1 + G2) v • v = [ 1 / (G1 + G2) ] i • i1 = G1 v • = [G1/ (G1 + G2) ] i • i2 = G2 v • = [G2/ (G1 + G2) ] i • The equations for i1 and i2 are “current divider” equations. The current i “divides” between the two resistances R1 and R2 in direct proportion to the sizes of the corresponding conductances G1 and G2. EEE 205 WS 2012 Part 4: Series & Parallel

  18. Current Division (2 resistors) Formula In Terms of R’s, [Instead of G’s] i2 + v – R1 R2 i1 i • i1 = G1/ (G1 +G2) ] i • = 1/R1 / [ (1/R1 + 1/R2) ] • = [ R2/ (R1 + R2) ] i • i2 = G2 / (G1 +G2) ] i • = 1/R2 / [ (1/R1 + 1/R2) ] i • = [ R1/ (R1 + R2) ] i • If R1 is large (relative to R2), then i1 is small. • If R2 is large (relative to R1), then i2 is small. • The larger current flows through the smaller resistor. EEE 205 WS 2012 Part 4: Series & Parallel

  19. Find i1 and i2 Example 4. i2 12  8 A i1 4  • i1 = [ 4 / (12 + 4) ] x 8 • = 2 A • The larger resistor has the smaller current. • i2 = [ 12 / (12 + 4) ] x 8 • = 6 A • The smaller resistor has the larger current. Solution: EEE 205 WS 2012 Part 4: Series & Parallel

  20. Current Division (N resistors) … Same voltage! + v – R1 R2 RN i1 i2 iN i … • i = (G1 + G2 + … + GN) v • v = [ 1 / (G1 + G2+ … + GN) ] i • For the kth resistor the current is: • ik = Gk v • = [Gk/ (G1 + G2+ … + GN) ] i • The current divides in proportion to the conductances. The larger currents flow through the larger conductances (smaller resistances). EEE 205 WS 2012 Part 4: Series & Parallel

  21. Find all the currents. Example 5. i1 7.6  52.5 A i2 15.2  i3 30.4  Solution: • i1 = 7.6–1 / ( 7.6–1 + 15.2–1+ 30.4–1) x 52.5 • = 30 A • i2 = 15.2–1 / ( 7.6–1 + 15.2–1+ 30.4–1 ) x 52.5 • = 15 A • i3 = 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) 2ndENTRY key to repeat the formula for editing. EEE 205 WS 2012 Part 4: Series & Parallel

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

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

  24. Find i and i1. Example 7 6  i1 4  4  6  i + – 1  12 V 10  3  Solution: 6  i1 4  6 + 6 = 12  4 // 12 = 3  4  6  i + – 1  12 V 10  3  3  4  4 + 3 + 3 = 10  10 // 10 = 5  i = 12 / 6 = 2 A i + – 1  12 V 10  3  EEE 205 WS 2012 Part 4: Series & Parallel

  25. Example 7. Solution (cont.) 10  1 A 12 V i = 2 A + – 1  10  1 A 6  i1 4  4  i = 2 A 1 A 6  + – 1  12 V 10  3  By current division: i1 = – 4–1 / (4–1 + 12–1 ) x 1 = – 0.75 A EEE 205 WS 2012 Part 4: Series & Parallel

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

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