Concept

Internal resistance — where it appears

The resistance in series with a source's own electromotive force, and the whole of what a load can discover about what is inside it. It is why a source model is a range rather than a component: the label stops describing the object at the current where that resistance is noticeable.

Named by 6 essays across 2 fields — each of them below, with the objects they name alongside it.

A 9 V source with 500 mΩ inside it. The ideal source is the flat line. The solved terminal voltage leaves it at a rate set entirely by the internal resistance: 1% low at 180 mA, half gone at 9.0 A.

The source that is not a source

An ideal voltage source holds its voltage at any current, which makes it the flattest line in the subject and the most commonly assumed model in it. Its edge is a current, set by one resistance nobody draws — and past that current the model is not approximately right, it is describing a different object.

networks · Source model

The equivalent of six elements, and the heat it does not account for. computed by solving, not by drawing at 73 loads. The network reduces to 8.56032 V behind 599.768 Ω, obtained twice and agreeing to twelve figures. Into every load on the axis the two put the same voltage to 2.2e-16 of it and deliver the same power to 7.8e-16 — there is no load that can tell them apart. Inside, at a load of 5.62 kΩ, the real network is dissipating 49.168 mW and the equivalent claims 1.1349 mW, a factor of 43.3. With the port open the equivalent says nothing at all is being burned and the network is burning 48.03 mW. The slider moves the resistor bridging the two sources, which joins two ideal sources and so cannot change anything the port can reach: the equivalent stays 8.56032 V behind 599.768 Ω to the last bit at every setting, while the heat inside moves by two and a half times.

Exact outside and wrong within

Six elements reduce to one source and one resistor that no load can distinguish from them: the same voltage into every load across six decades, to the last bit of a double. The reduction is wrong about the heat by a factor of forty-three, and with nothing connected it says the network is dissipating nothing while it burns 48 milliwatts.

networks · Equivalent circuit

The load that takes the most

A load equal to the source resistance takes more power than any other, and it does so at exactly fifty per cent efficiency — the source burns as much as the load receives. Ninety per cent efficiency needs a load nine times the source and delivers 36% of what was available, and a load half the source resistance delivers exactly as much as one twice it.

power · Power transfer

The load that takes the most power is 3.67 Ω, and the open-circuit voltage over the short-circuit current is 3.97 Ω. computed by solving, not by drawing. The terminal characteristic of a photocurrent with a junction across it, with the power along it drawn on the same axes and scaled to fill them. The most power, 80.99 W, is delivered at 17.244 V and 4.696 A, which is a load of 3.67 Ω. The incremental resistance of the source there — the negative of the characteristic's own slope — is 3.67 Ω, the same number to four figures. A straight line drawn between the two end points has a resistance of 3.97 Ω and would promise 24.81 W. The ratio of the true maximum to that promise is 3.264, and its reciprocal is the fill factor, 0.8160.

The load a curve recommends

The maximum power theorem says to match the load to the source's internal resistance, and for a straight-line source the internal resistance, the slope and the open-circuit voltage over the short-circuit current are one number. On a photovoltaic panel they are three: the slope at the maximum is 3.6717 ohms, the load there is 3.6717, and the open-circuit voltage over the short-circuit current is 3.9709 — and a quarter of the open-circuit voltage times the short-circuit current, which is what a straight line would deliver, is 24.81 watts against the 80.99 that is there. The theorem survives with the incremental resistance in place of the internal one, and it survives only where the characteristic has a slope: on a supply whose current limit folds back, the most power is delivered at a corner.

networks · Source model

One gain takes the equivalent resistance from 5 kΩ through infinity to negative. computed by solving, not by drawing. A 5 V source drives a node through 10 kΩ; the node also reaches 10 kΩ whose far end is held at A times the node's own voltage. The Thévenin resistance looking into that node is r1 in parallel with r2/(1 − A), which the solve returns to a part in a billion without being told: 5 kΩ at no gain, 10 kΩ at unity where r2 takes no current at all, and unbounded at A = 2.00 where the two conductances cancel. Above that it is negative. The open-circuit voltage follows it, because the short-circuit current is 500.0 µA at every gain — a short across the controlling node leaves the dependent source nothing to be controlled by — so the open-circuit voltage is simply the short-circuit current times whatever the resistance is, and reaches 150.0 volts from a five-volt source inside the range drawn.

The resistor that is not made of the resistors

Exact outside and wrong within reduced six elements to one source and one resistor and found the resistor two ways that agreed to the last bit. Put a dependent source in the network and one of those routes stops working, because setting the sources dead kills the independent ones and leaves the dependent one where it is. On a bootstrap of two ten-kilohm resistors the Thévenin resistance runs from five kilohms through infinity to minus ten, the open-circuit voltage of a five-volt source reaches 225, and above one gain the equivalent's resistor is negative — which the netlist refuses to stamp, correctly, because a negative resistance is a controlled source and not a resistor.

networks · Equivalent circuit $The load that may be complex, and what a resistor alone gives up. computed by solving, not by drawing, with the best load searched over BOTH of its parts on the solved network rather than substituted. Against a source of 50 Ω + j100 Ω the search returns 50.00 − j100.0 Ω — the conjugate — which delivers the available 500.00 mW at exactly 50.00% efficiency, and does so at every source reactance. A load that may only be a resistance takes 2/(1 + √(1 + x²)) of that, matching the solved answer to 4.4e-16 at ten reactances — and at x = 2 that fraction is exactly the golden ratio less one, 0.618034. The resistor-only load is the MORE efficient of the two at every non-zero reactance, rising towards one while the conjugate match sits at a half for ever, so the familiar "maximum power at fifty per cent" belongs to the conjugate and not to the load.$

The load that may be complex

Freed of the constraint that it be a resistance, the best load is the source's conjugate — found here by a two-dimensional search on the solved network rather than assumed — and it takes the available power at exactly fifty per cent efficiency whatever the source's reactance. A load that may only be a resistance takes 2/(1 + √(1+x²)) of that, and at a source reactance of twice its resistance that is exactly the golden ratio less one, 0.618034. The resistor-only load is also the MORE efficient of the two, rising towards one while the conjugate sits at a half for ever.

power · Power transfer

Named alongside it

The objects these essays reach for when they reach for this one.

Maximum power transfer Thevenin equivalent Design tradeoff Model range Impedance matching Output impedance Reactive power Short circuit current Verification Available power Bootstrapping Dynamic resistance

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