Reply To: understanding equivalent resistance in a parallel circuit

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Susan Brown

    Hi Michael,

    One way to think about it is to imagine things from an electron’s point of view. When it’s in a circuit, it can’t “see” how many loads there are and what their individual resistances are, it just “feels” (for lack of a better word) the total, or equivalent, effect of those loads.

    Kind of like when you are driving into a city and traffic suddenly slows down. You don’t know if there’s one big wreck ahead slowing things down, or a series of small incidents and distractions, or several different roads branching off that are backed up in various ways causing the overall slow-moving traffic. You are just sensing the net effect of whatever is up ahead.

    When we’re talking about two or more loads that are in series with each other, then we might need to know the “total resistance” of that circuit. For example, to determine the circuit current. In this case, we would just add up the resistances of all the loads. This makes sense, because the electrons in a series circuit only have that one path to flow in, and are affected by all the loads in that circuit.

    When we’ve got parallel branches, however, things are a little more complicated, because at some point along the journey from L1 to N (or L1 to L2), electrons have alternate routes they can take. But it can be handy to boil down those alternate, parallel paths, each with its own resistance, into one imaginary path with an “equivalent” resistance – the overall, or net, effect of those parallel paths.

    What can throw people is that the equivalent resistance of loads in parallel is *less* than any of the individual resistances of those loads.

    The thing to remember is that this is all from the perspective of the power supply. Be Zen-like and become the power supply, which has constant voltage, and think about the effect that having multiple paths for current to flow (parallel circuits vs. just one series circuit) has on the number of electrons you can push out (current). You can push a lot more out when there are multiple branches, even though each branch has a load.

    The best way to think this through is to draw out a couple of circuits on paper and play with the calculations.

    The first one has two loads in parallel, R1 and R2. Assume a 120vac circuit, so L1 and N. Assign easy values to R1 and R2, say 10 ohms and 20 ohms. Then calculate the current flowing through each load. (I = E/R, and remember in parallel circuits each load gets the full 120 vac.) The total current draw from L1 would be the sum of those two different currents.

    Now, let’s think about the “equivalent resistance” of this scenario. The equivalent resistance is taking the resistance of loads in parallel and theoretically combining them into a single load. So, do the math and come up with the equivalent resistance based on the loads above (we give you that formula in Fundamentals Mod3, unit 5). Now use I = E/R to calculate the current draw from L1. Should be the same as what you came up with before.

    Since current is inversely proportional to resistance, for the same amount of current to flow through our theoretical one load that was flowing through the two branches/loads in parallel, the single load, or equivalent load, would have to have a lower resistance.

    Again – try this yourself on paper. If the math gives you any challenges, let me know and I can help.

    Does this help at all? Let me know!