EE201 Lecture 5 P 1 Resistance Combinations Voltage and Current Division Resistors in series A vin t v1 v2 v3 C B R1 R2 iin t R3 R4 D v4 The circuit above can be represented by a single equivalent resistor and a voltage source veq vin t Req iin t What is Req EE201 Lecture 5 P 2 KCL 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 iin vj Rj iin Now apply KVL vin v1 v2 v3 v4 vj vin R1 R2 R3 R4 iin Rj iin Therefore Req R1 R2 R3 R4 Rk In series circuits Resistances add EE201 Lecture 5 P 3 Voltage division in series circuits Strategy express iin in terms of vin and Req using Ohm s law iin vin Req Substitute into vj Rj iin vj 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 4 Resistors in parallel iin t vin t R1 R2 R3 i1 i2 i3 The circuit above can be represented by a single equivalent resistor and a current source iin t veq Req What is Req EE201 Lecture 5 P 5 KVL 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 vin ij 1 Rj vin Gj vin Ohm s law Now apply KCL iin i1 i2 i3 ik iin G1 G2 G3 vin Gk vin Therefore Req G1 G2 G3 1 Gk 1 In parallel circuits Conductances add EE201 Lecture 5 Expressed in a more familiar way 1 Req 1 R1 1 R2 1 R3 P 6 1 1 Rk Current division in parallel circuits Strategy 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 resistors EE201 Lecture 5 P 7 Parallel series circuits Circuits with combinations of parallel and series subcircuits These can be reduced to equivalent circuits Example Find Req B A vin t 15 60 30 60 10 20 Solution 1 Find equivalent resistance of parallel subcircuit B Geq B 1 60 1 60 2 60 mhos Req B 1 Geq 30 EE201 Lecture 5 P 8 2 Find equivalent resistance of parallel subcircuit A Geq A 1 15 1 30 3 30 mhos Req 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 0 5 Iin Vin Vx 2 2 Iy 30 6 5 EE201 Lecture 5 P 9 Solution Branches 1 0 5 Iin Vin 3 2 Vx 2 Iy 30 6 2 5 Conductances 0 5 2 5 0 4 2 10 0 1 Resistances Req 1 Geq 1 Vin 12 V Vx 2 4 V Iy 1 A
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