(E-6) More Resistor NetworksThe 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.
An array of resistors in parallel
R_{1}, R_{2}, R_{3},...
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
So the total current is So, if R is the equivalent resistance of the circuit If as in the preceding section 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:
(2) Resistors of 60Ω, 90Ω and 150Ω in parallel ... R = ? ApplicationsHome 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 R_{4}
So R_{4} = 4Ω and the equivalent resistance of the entire circuit is 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 R_{i} of the conducting liquid ("electrolyte") which closes the circuit inside the battery or cell (to be sure, this is a simplified pictures). R_{i} 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 R_{3}=4Ω the internal resistance of the battery (actual value probably much less), and R_{1}=20Ω the resistance of the headlights, which draw a moderate current. Some voltage may drop while overcoming R_{3}, but not much. However, when the starter engine is connected (represented by R_{2}=5Ω) the effective load with R_{i}=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)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 R_{n}, its current I_{n} and the voltage difference ΔV_{n} between its ends (The Greek delta Δ, capital D in Greek, generally denotes "difference") satisfy Ohm's law 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 V_{1}, V_{2}, V_{3}, ... V_{6}. The first 3 voltages will be assigned to points (A_{1},A_{2},A_{3}) connected directly to A, the last 3 will be the voltages of (B_{1},B_{2},B_{3}) 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 V_{6} in terms of the others, use the relation to replace V_{6} in the remaining 5 equations. Then do the same to V_{5} 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 (A_{1},A_{2},A_{3})--in each the current has two exit routes towards B, and they look exactly alike. (Imagine twirling the cube around AB so that A_{2} replaces A_{1}, A_{3} replaces A_{2} and A_{1} replaces A_{w3}: the circuit still looks the same). We may therefore assign to them the same voltage V_{1} to these 3 points. Node B has exactly the same symmetry with respect to (B_{1},B_{2},B_{3}), 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 V_{2} to all 3 nodes.
Three currents emerge from A, each carrying 1/3 of the total current I and satisfying
(from here omit the Ω symbol). The current from A_{1} splits symmetrically into two equal parts, each and the same with A_{2} and A_{3}. Finally, three equal currents again unite at B, each one third of the total current: One can also omit the division by 1. Adding all three equations gives If R is the equivalent resistance of the cube, Ohms law gives If all resistors were different, you would have had a lot more work! Puzzlers(1) What is R if we add one more 1 Ω resistor, between A_{2} and A_{3}?Answer In the cube just studied, A_{2} and A_{3} have both the same voltage, namely V_{1}. 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 (A_{1},A_{2},A_{3}) 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 (B_{1},B_{2},B_{3}) 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 R_{1} = 1/3 Ω
(2) Suppose the three 1 Ω resistors from A to (A_{1},A_{2},A_{3}) are replaced by 2 Ω resisitors. What is R now? Equation (1) above now changes to
Multiply by 2 The other equations stay the same. Add as before The equivalent resistance is therefore 1 1/6 Ω or 7/6 Ω. Using the shortcut described above, R_{1} = 2/3 Ω and the same result is obtained. |
Home Page "All Things Electric and Magnetic"
"From Stargazers to Starships"
Timeline of "All Things Electric and Magnetic"
Timeline (from "Stargazers")
Master index file
Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last updated: 13 March 2010
Curators: Robert Candey, Alex Young, Tamara Kovalick
NASA Privacy, Security, Notices