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Kirchoff’s Laws and Equivalent Resistance Kirchoff’s Current Law At any junction point in an electrical circuit, the total current into the junction equals the total current out of the junction. Kirchoff’s Current Law At any junction point in an electrical circuit, the total current into the junction equals the total current out of the junction. (“What goes in must come out.”) In the diagram at right, I1 + I2 = I3 Kirchoff’s Voltage Law In any complete path in an electrical circuit, the sum of the potential increases equals the sum of the potential drops. Kirchoff’s Voltage Law In any complete path in an electrical circuit, the sum of the voltage increases equals the sum of the voltage drops. (“What goes up must come down.”) The Laws for a Series Circuit The current is the same at all points in the circuit: IT = I 1 = I 2 = . . . The total voltage supplied to the circuit is equal to the sum of the voltage drops across the individual loads: VT = V1 + V2 + . . . Ohm’s Law Equivalent Resistance in Series Equivalent (Total) Resistance in series: RT = R1 + R2 + R3…RN Equivalent (Total) Resistance in parallel: R T = 1 / (1/R1 + 1/R2 + 1/R3…1/RN) Equivalent Resistance in Parallel Equivalent (Total) Resistance in parallel: R T = 1 / (1/R1 + 1/R2 + 1/R3…1/RN) Equivalent Capacitance in Series Equivalent (Total) Capacitance in parallel: CT = C1 + C2 + C3…CN Equivalent Capacitance in Parallel Equivalent (Total) Capacitance in series: C T = 1 / (1/C1 + 1/C2 + 1/C3…1/CN) Series Resistance Example RT = I= Vdrop R1 = Vdrop R2 = Vdrop R3 = RT = 40 I = V/R = 60/40 = 1.5 A Vdrop R1 = 60 x (17/40) = 25.5 V Vdrop R2 = 60 x (12/40) = 18 V Vdrop R3 = 60 x (11/40) = 16.5 V 25.5 + 18 + 16.5 = 60 Vtotal Series Resistance Example RT = I= Vdrop R1 = Vdrop R2 = Vdrop R3 = RT = 40 I = V/R = 60/40 = 1.5 A Vdrop @ R1 = 60 x (17/40) = 25.5 V Vdrop @ R2 = 60 x (12/40) = 18 V Vdrop @ R3 = 60 x (11/40) = 16.5 V 25.5 + 18 + 16.5 = 60 Vtotal Series Resistance Example If you are more comfortable using Ohm’s Law to solve the Vdrop over each load, that will work too! V1 = I1R1 = (1.5 A)(17 W) = 25.5 V V2 = I2R2 = (1.5 A)(12 W) = 18 V V3 = I3R3 = (1.5 A)(11 W) = 16.5 V The Laws for a Parallel Circuit At a junction: IT = I 1 + I 2 + . . . But the total voltage across each of the branches is the same: VT = V1 = V2 = . . . Equivalent Resistance in Parallel Req = 1/ (1/R1 + 1/R2 + 1/R3 ) Req = 1/ (1/12 + 1/12 + 1/12 ) Req = 1/ 0.083 + 0.083 + 0.083 Req = 1/ .249 Req = 4.02 OR Req = 1/ (1/12 + 1/12 + 1/12 ) Req = 1/(3/12) Req = 1 / 0.25 Req = 4.0 Equivalent Resistance in Parallel Req = 1/ (1/R1 + 1/R2 + 1/R3 ) Req = 1/ (1/5 + 1/7 + 1/12 ) Req = 1/ 0.2 + 0.143 + 0.083 Req = 1/ 0.426 Req = 2.35 Combination Circuits What do we do if a circuit has both series and parallel loads? Find the equivalent resistance of the loads in parallel and continue the analysis. E.g.: Combination Circuits So…How would we find the current in this complex (combination) circuit?? Req = 1/ (1/R1 + 1/R2 + 1/R3 ) Req = 1/ (1/4 W + 1/4 W ) Req = 1/ (2/4) Req = 1/ (.5) Req = 2 First, find the resistance in the parallel branch. Combination Circuits Req = R1 + Req2+3 + R4 Req = 2 + 2 + 6 Req = 10 Then find the overall resistance in the circuit. Combination Circuits I = V/R I = 12/10 I = 1.2 A Then use Ohm’s Law to find the current in the circuit. Combination Circuits What is the voltage drop here? What is the voltage drop here? I = 1.2 A Req = 10 V = 12 V What is the voltage drop here? (Across ENTIRE branch.) Current will split here into two (2) equal paths of 0.6 A each. Why equal parts? Now you should be able to describe many things about the circuit…. Combination Circuits V= I x R V = 1.2 x 2 V = 2.4 V= I x R V = 1.2 x 6 V = 7.2 Vtot = 2.4 + 2.4 + 7.2 Vtot = 12 V I = 1.2 A Req = 10 V = 12 V V= I x R V = 1.2 x 2 V = 2.4 If we measure the voltage drop across each of the individual 4 resistors inside the parallel branch, what will we find?