how you would measure the internal resistance of the power supply?

Question

You may assume that the multimeters are ideal.

Answer 1

In practice, I would connect a known resistance (conductance Go) representing somewhat lighter than the typical load on the supply, measure the voltage across it, Then an additional test load Gt in parallel with Go and measure the voltage again. I do this to allow for nonlinearities in the effective series resistance (or parallel conductance) of the power supply. If there is no reason to suspect a nonlinearity, I would of course let Go be zero, i.e., an open circuit, and get a more accurate reading that way. Let Vo be the first reading and Vt be the second reading, and let Gs be the source conductance. If we now model our source as a ideal current (!) source of Ig, we get the equations

Ig = Vo (Gs + Go) = Vt(Gs + Go + Gt)

What we really want is Gs, which we can invert to get Rs, so we solve the equation
Vo(Gs + Go) = Vt(Gs + Go) + VtGt
to get Gs = VtGt / (Vo-Vt) - Go

Answer 2

1. Measure the output of the supply under no load. (V0)
2. Apply a known load (I amps) and measure the voltage (Vl)
3. Calculate (V0 - Vl)/I and that's the Thevenin equivalent internal resistance of the supply.

(You can do the same with a reactive load ant a known frequency to get the reactive part of the internal impedance, but it's usually so low that nobody ever does it)

Doug

Answer 3

An ideal voltage source would have a resistance that can be calculated using ohms law.

Setup your source in series with your resistor then use the following formula.

Vsource = Current/(internal resistance + resistance)

Therefore your internal resistance would be:

Internal Resistance = (Current - Vsource* Resistance)/VSource

Answer 4

Just hook up an ohm meter to the power source.

Answer 5

By Meter of its....
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