Opening Kirchoff! Kirchoff’s Laws apply to branched and multi-loop circuits. There are two of them, the first states that the sum of the currents coming into a junction equals the sum of the currents leaving the junction (charge conservation, no build-up of charge at a junction. The second states that the sum of the voltage differences around any closed loop is equal to zero (potential has one value at any given point). Consider the following circuit with a 10V source and all resistors = 1kΩ. We wish to determine the current through each resistor and the voltage at points a , b and c. This circuit can be analyzed either by series and parallel combinations or by Kirchoff’s Laws. We will use Kirchoff’s Laws and then check it with the series and parallel combinations. Current splits: I1 = I2+I3+I4 I5 = I3+I4 I5+I2 = I1 Ioops: (clockwise) L1 : V - I1R1 – I2R2 = 0 L2: I2R2 – I3R3 – I5R5 = 0 I2 = I3+I5 = 1.5 I5 using the line below. L3: I3R3 – I4R4 = 0 I3 = I4 = ½ I5 using I5 = I3+I4 Using I1 = I5+I2, = I5 + 1.5 I5 or I1 = 2.5 I5 rewrite all currents in terms of I5: I1 = 2.5 I5 I2 = 1.5 I5 I3 = I4 = ½ I5 L1 : V - I1R1 – I2R2 = 0 = V -1kΩ(2.5 I5 + 1.5 I5) = V – 4 (I5) (1kΩ) = 0 From L1 with V = 10V, 10V - (I5) (4kΩ) = 0 I5 = 2.5 mA Using the current relations, this gives: I1 = 2.5 I5 = 6.25mA I2 = 1.5 I5 = 3.75mA I3 = I4 = ½ I5 = 1.25mA Voltages: c = 0V, no resistance between c and battery. a = 10V – I1R1 = 10v – (6.25mA)(1kΩ.) = 3.75V b = 0V + I5R5 = (2.5mA)( 1kΩ.) = 2.5V Check the results using series and parallel combinations to find total resistance and current from battery I1. R3 and R4 are in parallel. The resulting R is in series with R5. This combination is in parallel with R2. All of that is in series with R1. Calculate the result using 1kΩ resistors: R3 ll R4 : 1/1kΩ + 1/1kΩ = 1/R = 2/1kΩ or R = 0.5 kΩ R + R5 = 1.5 kΩ R2 ll 1.5 kΩ = 1/1kΩ + 1/1.5kΩ = 5 / 3kΩ so R = 0.6kΩ R1 + 0.6kΩ = 1.6kΩ = Rtot 10V = Itot Rtot = Itot (1.6kΩ) Itot = 10V / Rtot = 10V / 1.6kΩ = 6.25mA same as before. The rest follows from the current splits and the equivalent resistances. After R1, the current splits between 2 branches with R2 = 1kΩ and the combination R = 1.5 kΩ Since the branches have the same voltage difference across them, the IR products must be equal. So I2 = (1.5)I5 and I2+I5 = I1 = 6.25mA = (2.5)I5 giving I5 = 2.5mA and I2 = 3.75mA Finally, I3 = I4 = ½ I5 = 1.25mA All of which agrees with the previous method. The voltages are calculated using Ohm’s law as before. Some circuits cannot be fully analyzed using series and parallel formulas. For these, Kirchoff’s Laws are the tools to use. Consider the following circuit. Unless there are equivalent resistors in the split diamond configuration, series and parallel formulas cannot be used to break down the combination to equivalent resistances. The structure of the circuit is more complex than just series and parallel. For this example, let V = 10 V and the resistors Rx = x kΩ so they are all different. The directions for the currents are chosen arbitrarily; if they are reversed, the algebra will let us know with a negative sign. The copy of the diagram below has the currents and loops labeled. The current rule tells us that a. I1 = I2 + I3 b. I2 = I4 + I5 c. I4 + I3 = I6 I5 + I6 = I1 = total current. The loop rule sums ∆V’s around a loop. Ohm’s law gives us ∆V=IR for a resistor. This is negative (a drop in V) in the direction of the current and positive when going opposite the current (up hill). For loop L1, this gives V – I1R1 – I2R2 – I5R5 = 0 L2: +I5R5 – I4R4 –I6R6 = 0 L3: - I2R2 – I5R5 + I6R6 + I3R3 = 0 a. I1 = I2 + I3 L1: V – I1R1 – I2R2 – I5R5 = 0 b. I2 = I4 + I5 L2: +I5R5 – I4R4 –I6R6 = 0 c. I4 + I3 = I6 L3: - I2R2 – I5R5 + I6R6 + I3R3 = 0 I5 + I6 = I1 = total current. First eliminate some of the currents: I3 = I1 – I2 I6 = I1 – I5 I4 = I2 – I5 Now substitute these into the equations for loops L2 and L3: L2: +I5R5 – (I2 – I5)R4 –(I1 – I5)R6 = 0 L3: - I2R2 – I5R5 + (I1 – I5)R6 + (I1 – I2)R3 = 0 Rewrite, L2: +I5R5 – I2R4 + I5R4 –I1R6 + I5R6 = 0 I5(R5+R4+R6) –I2R4 – I1R6 = 0 I5(R5+R4+R6) = I2R4 + I1R6 or I5 = (I2R4 + I1R6) / (R5+R4+R6) L3: - I2R2 – I5R5 + I1R6 – I5R6 + I1R3 – I2R3 = 0 - I5(R5+R6) – I2(R2+R3) + I1(R3+R6) = 0 or [- (R5+R6) / (R5+R4+R6)] (I2R4 + I1R6) – I2(R2+R3) + I1(R3+R6) = 0 I1 [(R3+R6) –R6(R5+R6) / (R5+R4+R6)] = I2 [(R2+R3) + R4(R5+R6) / (R5+R4+R6)] I1 [(R3R5+R3R4+R3R6+R4R6) / (R5+R4+R6)] = I2[{(R2+R3)(R5+R4+R6) + R4(R5+R6)} / (R5+R4+R6)] I1 [(R3R5+R3R4+R3R6+R4R6)] = I2 [(R2+R3)(R5+R4+R6) + R4(R5+R6)] I2 = I1 {[(R3R5+R3R4+R3R6+R4R6)] / [(R2+R3)(R5+R4+R6) + R4(R5+R6)]} = I1 { } for short. Return to I5: I5 = (I2R4 + I1R6) / (R5+R4+R6) I5 = (I1{ }R4 + I1R6) / (R5+R4+R6) I5 = I1 (R4{ }+R6) / (R5+R4+R6) L1: V – I1R1 – I2R2 – I5R5 = 0 V = I1R1 + I2R2 + I5R5 V = I1R1 + I1 { }R2 + I1 R5(R4{ }+R6) / (R5+R4+R6) = I1 [R1 + { }R2 + R5(R4{ }+R6) / (R5+R4+R6) ] So, with { } = {[(R3R5+R3R4+R3R6+R4R6)] / [(R2+R3)(R5+R4+R6) + R4(R5+R6)]} I1 = V / [R1 + { }R2 + ( R5(R4{ }+R6) / (R5+R4+R6) ) ] I5 = I1 (R4{ }+R6) / (R5+R4+R6) I2 = I1 { } I3 = I1 – I2 I6 = I1 – I5 I4 = I2 – I5 From these, Ohm’s Law ΔV = IR can be used to calculate voltages. With numbers for all the resistance values, the expressions all boil down to numbers and don’t look so terrifying. Try calculating the currents and voltages at all junctions using [1] R1 = 0, R2=R3=2kΩ, R4=100Ω, R5=R6=1kΩ (What is I4 and what are the voltages at b and c?) [2] R1=0, R2=R6=2kΩ, R4=100Ω, R5=R3=1kΩ