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(E-6)   More Resistor Networks

Resistors connected in series act like a single resistor of equivalent resistivity R equal to the sum of all the individual resistors
R = R1 + R2 + R3 + R4 + ...

The same current I flows through all of them, but the imposed voltage V is divided among the resistors in proportion to their resistance--more of it drops across the larger resistance, less across the smaller one.

    The other "standard" connection scheme of resistors is in parallel--all resistors receiving input from the same point, and supplying their current to the same point.

In this case, all resistors sense the same voltage V, but the current I is divided between them, the smaller resistance drawing more current.

    (By the way, electric cells may also be connected in series or in parallel. Connecting them in series produces the usual electric battery, whose voltage V is the sum of the voltages of all its cells. Connecting them in parallel the voltage is the same as that of one cell, but a much larger current can be extracted. However, all cells better have the same voltage here, or those of higher voltage will start charging those of lower voltage.)

    An array of resistors in parallel   R1, R2, R3,... may also be replaced by "an equivalent resistance" R. Suppose we have 3 resistors. Since all are fed from the same voltage V and Ohm's law is obeyed by each, theur currents are

I1 = V / R1             I2 = V / R2             I3 = V / R3

So the total current is
I  =   I1 + I2 + I3   =   V/R1 + V/R2 + V/R3  =   V {1/R1 + 1/R2 + 1/R3}

So, if R is the equivalent resistance of the circuit

  I =   V / R
1/R   =   1/R1 + 1/R2 + 1/R3

If as in the preceding section

R1 = 10Ω         R2 = 20Ω         R3 = 25Ω        
1/R  =   0.1 + 0.05 + 0.04  =   0.19 = 19/100
R  =   100/19 Ω = 5.263 Ω

The rule is: resistances in series have a resistance larger than any of them, while resistances in parallel have a resistance smaller than any of them.

Practice problems to work out:

    (1) 5 resistors of 10 Ω in parallel... R = ?
    (2) Resistors of 60Ω, 90Ω and 150Ω in parallel ... R = ?
(answers--two ohms and about twenty nine ohms)


    Home circuits are always wired with all outlets and connections in parallel, because each of them must be fed from the same 110v supply (220v in Europe). Suppose you wired a room for two identical lightbulbs, and when you turned on the circuit, both switched on together, as they were supposed to--but their light was dim and their filaments glowed weakly. Why?

    Most probably because you wired both lamps in series, not in parallel, so that the same current passes through both and each gets only 55 volts. Want to be sure? Unscrew one--the other should go out, too.

    Of course, the total current you can draw from a home circuit is limited. If you connect too many lights and appliances, or somehow connect wires with opposite voltages directly (a "short circuit" with very little resistance), your circuit will draw so much power that your wiring gets dangerously hot. In most cases, a fuse will blow or a circuit breaker (to be discussed later) will cut the power, and the circuit will go dark. That will be your warning--before an excessive current melts your wiring, burns insulation or perhaps starts a house fire. Newspaper reports of fires often end with "...the fire chief believes the fire had an electrical origin": that is how it happens.

    To derive the current drawn by a more complicated circuit, you can often start by replacing combinations of resistors with equivalent resistances. For instance, with the circuit shown here you first replace the two resistors in parallel with their equivalent single resistance R4

1/R4   =   1/20 + 1/5   =   0.05 + 0.2   =   0.25   =   1/4

So R4 = 4Ω and the equivalent resistance of the entire circuit is

R   =  R3 + R4  =   8 Ω

    Such a situation arises with batteries and cells under heavy load. The current they supply has to overcome not only the resistance, in the external circuit but also the internal resistance Ri of the conducting liquid ("electrolyte") which closes the circuit inside the battery or cell (to be sure, this is a simplified pictures). Ri is generally small, but it sets a limit to the current which can be extracted, even when the terminals are connected directly ("shorted out").

    The above circuit (conceptually, not in actual values) may be viewed as representing the circuit of a car, with R3=4Ω the internal resistance of the battery (actual value probably much less), and R1=20Ω the resistance of the headlights, which draw a moderate current. Some voltage may drop while overcoming R3, but not much. However, when the starter engine is connected (represented by R2=5Ω) the effective load with Ri=4Ω equals the internal resistance of the battery, and therefore only half the available voltage appears across the circuit AB of the headlights, and those lights dim briefly, as is often observed. To test yourself, you may try derive the equivalent resistance of the circuit below (we get two hundred fifty one point twenty eight, approximately)

Node analysis             (optional)

(it is recommended that you follow these calculations on a sheet of paper)

    Not all resistor circuits can be resolved into combinations of resistors in parallel and in series. Consider a "classical" problem appearing in many textbooks: A 10V voltage is applied between opposite corners (A,B) of a cube consisting of 12 resistors of 1Ω each. How much current will flow--or in other words, what is the equivalent resistance R between A and B?

This circuit cannot be resolved into separate resistors in parallel and in series (at least not in the usual way). However, we know that for any nth resistor Rn, its current In and the voltage difference ΔVn between its ends (The Greek delta Δ, capital D in Greek, generally denotes "difference") satisfy Ohm's law

In   =   ΔVn/ Rn

As long as that law holds, the resistance of an array is the same for any applied voltage, so let us assume ΔV = 10V between A and B--say V=10 at A and V=0 at B, since the choice of the point where V=0 is arbitrary. That leaves 6 corners of "nodes" where wires meet, with voltages V1, V2, V3, ... V6. The first 3 voltages will be assigned to points (A1,A2,A3) connected directly to A, the last 3 will be the voltages of (B1,B2,B3) connected directly to B

    There also exist 6 equations, one for each node, expressing the requirement that "what flows in equals what flows out". Six equations with 6 unknown can usually be solved: use one to express V6 in terms of the others, use the relation to replace V6 in the remaining 5 equations. Then do the same to V5 reducing the number of equations to 4, and so on, until one voltage remains, expressed by the remaining one equation.

    It is doable, and if all resistors were different, this sort of drudgery might be the only way (though a computer code may help). Here however every resistor the same, namely 1Ω, and one can take advantage of symmetries.

    If you look at the cube in the direction of the 3-dimensional diagonal AB, you will notice a 3-fold symmetry in the position of the nodes (A1,A2,A3)--in each the current has two exit routes towards B, and they look exactly alike. (Imagine twirling the cube around AB so that A2 replaces A1, A3 replaces A2 and A1 replaces Aw3: the circuit still looks the same). We may therefore assign to them the same voltage V1 to these 3 points.

    Node B has exactly the same symmetry with respect to (B1,B2,B3), each receives currents from A by two identical routes, and twirling the cube just replaces each one by another. We can therefore again assign here the same voltage V2 to all 3 nodes.

    Three currents emerge from A, each carrying 1/3 of the total current I and satisfying

I/3   =   (V – V1)/ 1Ω            (1)

(from here omit the Ω symbol). The current from A1 splits symmetrically into two equal parts, each
I/6   =   (V1 – V2)/ 1            (2)

and the same with A2 and A3. Finally, three equal currents again unite at B, each one third of the total current:
I/3   =   (V2 – 0)/ 1            (3)

One can also omit the division by 1. Adding all three equations gives

I(1/3 + 1/6 + 1/3)   =   (V – V1)+ (V1 – V2) + V2
(5/6) I   =   V

If R is the equivalent resistance of the cube, Ohms law gives

V = I R             R = 5/6 Ω

If all resistors were different, you would have had a lot more work!


(1) What is R if we add one more 1 Ω resistor, between A2 and A3?

    Answer     In the cube just studied, A2 and A3 have both the same voltage, namely V1. If we connected them by resistor (of whatever value), no current would flow in it, because both ends are at the same voltage. So R remains the same, 5/6 Ω

    This suggests the problem could be solved even faster by a trick based on its symmetry. Since (A1,A2,A3) are all at the same voltage, we might as well connect them by a wire of negligible resistance, so in the electric circuit they become the same point--call it "C".

    Similarly (B1,B2,B3) are at the same voltage, may also be connected directly and be electrically the same point in the circuit--call it D. With these shortcuts

From A to C : three 1 Ω resistors in parallel, equivalent to R1 = 1/3 Ω
From C to D : six 1 Ω resistors in parallel, equivalent to R2 = 1/6 Ω
From D to B again three 1 Ω resistors in parallel, equivalent to R3 = 1/3 Ω

    The 3 equivalent resistors are connected in series, so they give

R = R3 + R6 + R3 = 5/6 Ω

(2) Suppose the three 1 Ω resistors from A to (A1,A2,A3) are replaced by 2 Ω resisitors. What is R now?

Equation (1) above now changes to

I/3   =   (V – V1)/ 2Ω            (1')

Multiply by 2
2 I/3   =   (V – V1)/ 1Ω            (1")

The other equations stay the same. Add as before

I(1/3 + 1/6 + 1/3)   =   V

I(1 1/6)   =   V

The equivalent resistance is therefore 1 1/6 Ω or 7/6 Ω. Using the shortcut described above, R1 = 2/3 Ω and the same result is obtained.


Back to E5: Resistor circuits--using Ohm's Law
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