SINGLE LOOP CIRCUITS Background: using kvl and kcl we can write enough equations to analyze any linear circuit. we now start the study of systematic, and efficient, ways of using the fundamental circuit laws a 1 2 b 6 branches 6 nodes 1 loop 3 c 4 VOLTAGE DIVISION: THE SIMPLEST CASE write 5 kcl eqs or determine the only current flowing KVL ON THIS LOOP f 6 e 5 d ALL ELEMENTS IN SERIES ONLY ONE CURRENT the plan • begin with the simplest one loop circuit • extend results to multiple source • and multiple resistors circuits Important voltage divider equations 1 ‧節點node:兩個以上的電路元件連接的地方。 ‧本質節點essential node:三個以上電路元件 連接的一種節點。 ‧路徑path:由基本元件連成,但每個元件只能 通過一次的路線。 ‧分支branch:連接二個節點的路徑。 ‧本質分支essential branch:連接二個本質節 點但不通過別的本質節點的路徑。 ‧迴路loop:終點跟起點是同一個節點的路徑。 ‧網目mesh:不再含其他迴路的一種迴路,也就 是最小的迴路。 2 SUMMARY OF BASIC VOLTAGE DIVIDER v R1 = R1 v (t ) R1 + R2 EXAMPLE : VS = 9V , R1 = 90kΩ, R2 = 30kΩ VOLUME CONTROL? R1 = 15kΩ ⇒ 3 A “PRACTICAL” POWER APPLICATION 4 THE CONCEPT OF EQUIVALENT CIRCUIT THE DIFFERENCE BETWEEN ELECTRIC CONNECTION AND PHYSICAL LAYOUT THIS CONCEPT WILL OFTEN BE USED TO SIMPLFY THE ANALYSIS OF CIRCUITS. WE INTRODUCE IT SOMETIMES, FOR PRACTICAL CONSTRUCTION HERE WITH A VERY SIMPLE VOLTAGE DIVIDER REASONS, COMPONENTS THAT ARE ELECTRICALLY CONNECTED MAY BE PHYSICALLY QUITE APART i vS R1 i vS + - R2 i= R1 + R2 + - vS R1 + R2 AS FAR AS THE CURRENT IS CONCERNED BOTH CIRCUITS ARE EQUIVALENT. THE ONE ON THE RIGHT HAS ONLY ONE RESISTOR SERIES COMBINATION OF RESISTORS R1 R2 ≡ R1 + R2 IN ALL CASES THE RESISTORS ARE CONNECTED IN SERIES 5 FIRST GENERALIZATION: MULTIPLE SOURCES + v2 − + v R1 − Voltage sources in series can be algebraically added to form an − equivalent source. v3 + - v1 + - − v5 i(t) R2 + - − R1 + - + + + We select the reference direction to + move along the path. v Voltage drops are subtracted from rises R2 − + - KVL − v4 + vR1 + v2 − v3 + vR 2 + v4 + v5 − v1 = 0 R1 Collect all sources on one side (v1 − v2 + v3 − v4 − v5 ) = vR1 + vR 2 (v ) = v eq R1 + vR 2 veq + - R2 6 SECOND GENERALIZATION: MULTIPLE RESISTORS FIND I ,Vbd , P (30k ) APPLY KVL TO THIS LOOP APPLY KVL TO THIS LOOP LOOP FOR Vbd Vbd − 12 − 20 [k Ω] I = 0 (KVL) ⇒ Vbd = 10V POWER ON 30k Ω RESISTOR P = I 2 R = (−10 −4 A) 2 (30 *103 Ω) = 30mW v R = Ri i ⇒ i VOLTAGE DIVISION FOR MULTIPLE RESISTORS 7 THE “INVERSE” VOLTAGE DIVIDER R1 + VS + - R2 VO − VOLTAGE DIVIDER VO = R2 VS R1 + R2 "INVERSE" DIVIDER VS = R1 + R2 VO R2 COMPUTE VS " INVERSE" DIVIDER VS = 220 + 20 458.3 = 500kΩ 220 8 Find I and Vbd APPLY KVL TO THIS LOOP − 6 + 80kI + 12 + 40kI = 0 ⇒ I = −0.05mA Vbd − 40kI − 12V = 0 ⇒ Vbd = 10V 9 If Vad = 3V, find VS + 3V − INVERSE DIVIDER PROBLEM 25 + 15 + 20 VS = 3 = 9V 20 10 Notice use of passive sign convention + 80k Ω * i ( t ) − KVL : − 6V + 80k Ω * i ( t ) + 12V + 40k Ω * i ( t ) = 0 + i(t ) i(t ) = − 6V = −0.05mA 120k Ω 40k Ω * i ( t ) − Knowing the current one can compute ALL the remaining voltages and powers 11 EXAMPLE A 9V 20k B + - C I + - KVL FOR E 30k DETERMINE I VDAUSING KVL D 10k V DA = VCD = 30k * I = 1.5V I DE = 0.05mA KVL : - 12 + 20k * I + 9 + 30k * I + 10k * I = 0 3V I= = 0.05mA 60kΩ KVL : V DA + 12 − 10k * I = 0 V DA = −11.5V 12 − Vab + KVL HERE EXAMPLE + 4V − b +- 4k VS Sometimes you may want to vary a bit 3Vx + - a I APPLY KVL TO THIS LOOP VS = 12V + Vx V X Vab = = − P(3Vx ) = OR KVL HERE P( 3Vx ) is the power absorbed or supplied by the dependent source KVL : − 12 + 4 + 3V X + V X = 0 ⇒ V X = 2V KVL : Vab + 4 + 3V X = 0 ⇒ Vab = −10V KVL : Vab + VS − V X = 0 P(3V ) = 3V X I (PASSIVE SIGN CONVENTION) X 4V = 1mA 4k = 2[V ] *1[mA ] = 2mW OHMS' LAW : I = P(3V X ) 13