Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9EE201 Lecture 5 P. 1Resistance Combinations, Voltage and Current DivisionCBDA+_Resistors in series_+_+_+_+R1R2R3R4v1v2v3v4iin(t)+_veq_+ReqThe circuit above can be represented by a single equivalent resistor and a voltage source. vin(t) vin(t)iin(t)What is Req?EE201 Lecture 5 P. 2KCL: current is constant in a series circuit. KVL: sum of voltages around a closed loop is zero.Strategy: express voltage drop for each resistor in terms of resistance at time t (subscripts dropped). v1 = R1 iin; v2 = R2 iin ; v3 = R3 iin ; v4 = R4 iinvj = Rj iinNow apply KVL,vin = v1 + v2 + v3 + v4 = vjvin = (R1 + R2 + R3 + R4) iin = ( Rj) iinTherefore, Req = (R1 + R2 + R3 + R4) = Rk In series circuits: Resistances addEE201 Lecture 5 P. 3Voltage division in series circuitsStrategy: express iin in terms of vin and Req using Ohm’s law. iin = vin / ReqSubstitute into, vj = Rj iinvj = Rj (vin / Req) = (Rj / Rk) vin e.g. R3 v3 = (vin)(R1 + R2 + R3 + R4) Voltage division equation for resistors in series.EE201 Lecture 5 P. 4Resistors in parallelveq_+ReqThe circuit above can be represented by a single equivalent resistor and a current source. vin(t)R1R2R3_+i1i2i3 iin(t) iin(t)What is Req?EE201 Lecture 5 P. 5KVL: voltage drop across all resistors are equivalent. KCL: the sum of currents entering a node equals the sum of currents leaving a node.Strategy: express current for each resistor in terms of resistance and vin (t) at time t (subscripts dropped). i1 = (1/R1) vin; i2 = (1/R2) vin ; i3 = (1/R3) vinij = (1/Rj ) vin = (Gj ) vin(Ohm’s law)Now apply KCL,iin = i1 + i2 + i3 = ikiin = (G1 + G2 + G3) vin = ( Gk) vinTherefore, Req = (G1 + G2 + G3)-1 = ( Gk)-1In parallel circuits: Conductances addEE201 Lecture 5 P. 6 1 1Req = = (1/R1 ) + (1/R2 ) + (1/R3) (1/ Rk)Expressed in a more familiar wayCurrent division in parallel circuitsStrategy: express vin(t) in terms of iin(t) and Req using Ohm’s law. vin (t) = Req iin (t)Substitute into ij (t) = (1/Rj ) vin (t) = (Gj ) vin (t) ij (t) = (1/Rj ) Req iin (t) = (Req / Rj ) iin (t) ij (t) = Gj (1/Geq) iin (t) = (Gj / Geq ) iin (t) current division equation for parallel resistorsEE201 Lecture 5 P. 7Parallel-series circuitsCircuits with combinations of parallel and series subcircuits. These can be reduced to equivalent circuits. Example: Find Req.+_10 vin(t)15 30 60 60 20 ABSolution: 1) Find equivalent resistance of parallel subcircuit B. Geq(B) = (1/60 ) + (1/60 ) = (2/60) mhosReq (B) = 1 / Geq = 30 EE201 Lecture 5 P. 82) Find equivalent resistance of parallel subcircuit A. Geq(A) = (1/15 ) + (1/30 ) = (3/30) mhosReq (A) = 1 / Geq (A) = 10 3) Sum series resistances.Req = 10 + 10 + 30 + 20 = 70 Example: Find Req, Geq, Vin, Vx and Iy if Iin = 12 A. Vin2_+ Iin Vx_+2 5306 Iy0.5EE201 Lecture 5 P. 9Solution: Vin2_+ Iin Vx_+2 5306 Iy0.5123Req = 1/ Geq = 1 Vin = 12 VVx = 2.4 V Iy = 1 ABranches:2 2.510Resistances:0.5
View Full Document
Unlocking...