Beyond Series and Parallel

I think the simplest one-battery circuit that you cannot analyze as a combination of series and parallel is the one below. Given e and R₁ thru R₅, how do you find the battery current and the other currents and voltage drops? (There is an easy special case: if R₁/R₄ = R₂/R₅, it is not difficult to show that the V across R₃ is zero, hence removing it will have no effect, hence the total resistance is (R₁ + R₄) in parallel with (R₂ + R₅). But in general we cannot do this.) On this page we will explore some ways of tackling this problem. It will definitely be more fun than scrubbing out a latrine with a toothbrush.

The hard way to solve it (so skip these two paragraphs):
(This is Kirchoff's rules method. It is easy conceptually, but the math is tedious.) Let I_B = battery current, I₁ = current in R₁, etc. Write equations involving currents. I_B = I₁ + I₂ = I₄ + I₅, I₁ = I₃ + I₄, I₂ + I₃ = I₅. Note 1: It appears that I made the assumption that I₃ is directed downward, but really the direction does not matter. If it is upward, the math will give me the correct negative value. Note 2: Those current equations are not independent, which means from two of them you can deduce the third (try it). Let us choose the latter two to work with.

Now write voltage equations based on the principle that the sum of the voltage changes around any closed loop is zero. We need 3 equations to go with the 2 current equations so that we can find 5 unknown currents. Here are three that would do the job: e -I₁R₁ - I₄R₄ = 0, I₁R₁ - I₂R₂ + I₃R₃ = 0, e - I₁R₁ - I₃R₃ - I₅R₅ = 0. Now we have 5 equations. If e and all R's are known, we have 5 unknown currents. The best way to deal with this is with determinants and Cramer's rule. If you have not had that stuff in math, do it by substitution, eventually reducing down to one unknown. Or use a computer program such as Maple, Mathcad, Mathematica, Matlab, ….I quit. Let me show you two easier ways.

Easier way # 1, the D - l theorem
It can be shown that any three resistors connected in the shape of a D in a circuit can be replaced by three different resistors in the shape of a l (see sketch below), and the currents, etc. in the rest of the circuit are unchanged if we make R_A = R₁R₂/(R₁ + R₂ + R₃), R_B = R₂R₃/(R₁ + R₂ + R₃), and R_C = R₁R₃/(R₁ + R₂ + R₃). In the circuit at the top of the page, R₁, R₂ and R₃ are a D , so make the switcheroo, and we then have R_A in series with the parallel combination (R_B + R₄) in parallel with R_C + R₅, so now we can easily calculate everything.

Easier way # 2, Thevenin's theorem
You can find the current through any resistor connected between points A and B in a circuit (such as at the top of the page) if you can calculate the voltage that would be from A to B with R removed (V_AB), and the resistance from A to B of the rest of the circuit (R_CIRC ), again with R removed, and this time replacing batteries with their internal resistances. Then the current through R is what you get if you replace the rest of the circuit with a battery of voltage V_AB and resistor R_CIRC in series with it. Put R back in and we have I = V_AB/(R_CIRC +R). This is valid for any network of resistors and batteries.

Actually in this particular problem, you might find this method difficult, because both V_AB and R_CIRC are tricky to calculate. But in some problems, this method is a piece of cake. If you are seriously into this stuff, work out a problem or two by all methods.

The circuit below is designed to make R_CIRC a quantity that you can calculate in your head. It is a simple series-parallel combination, but it is tricky. If you can get it in two minutes, you are a genius. (If it takes an hour, you might still be a genius, but you got off on the wrong track.)

The convention is that 6 k means 6 kW , or 6000 W . Assume that the 288 volt power supply has zero internal resistance, so replace it with a wire when you are finding the R_CIRC (and that is the part that makes it tricky-- you might want to redraw the circuit). Remember that two resistors in parallel is equivalent to the product divided by the sum; also you can work it in terms of kW . Now find the current in the two resistors on the left, using Thevenin's theorem.

Do not read beyond this point until you have worked it out, then read below and see if you agree with my answer.

Remove the 6 k on the top of the circuit and find that R_CIRC is 2 k. Put the 6 k back and remove the 30 k, and find that this R_CIRC is also 2 k! Find those voltages, and show that I_6k is 34 mA and I_30k is 7 mA, so that I_B is 41 mA. Then use the D - l theorem to convert it to a series parallel combination, calculate its resistance, and verify that the battery current is 41 mA. Use any of various methods now and show that the current in the 1 k is 6 mA toward the bottom of the page, I_3k = 28 mA and I in the other 6k is 13 mA. Verify the Kirchoff's rules (the equations discussed first).

Flow back to the main page on electricity, etc., or send me hate mail, virus, etc. at fredrick.gram at tri-c.edu