Internal Resistance of Voltage Sources
How to find the internal resistance of a voltage source? The internal resistance can be find through a test under no-load and full load conditions. Under No-Load conditions, the supply on
The power dissipated by the internal resistance, represents the heat generated in the power supply. This is illustrated in the animation below. The terminal voltage (V) is equal to the e.m.f. voltage (E), minus the internal voltage drop (Ir). (using ohms law: internal voltage drop = current (I) x internal resistance (r)).
To measure DC internal resistance with a multimeter, you first measure the unloaded voltage of the battery (v1), then the voltage under load (v2), and finally the resistance of the load (r1), which allows you to calculate the internal resistance using ISR = (V1 - V2)/ (V2/R1).
To find the internal resistance of the voltage source channels, its just 30V/3A = 10Ohms and 5V/3A = 1.66Ohms. Is it correct? Eh. No. If that were the internal resistance, then a load of 3A would cause the output voltage to drop drastically. Ideal voltage sources have zero internal resistance - the current drawn is solely due to the external load.
The resistance level depends on the type of supply, but it is always present. Every year new supplies come out that are less sensitive to the load applied, but even so, some sensitivity still remains. Due to the internal resistance of the supply, the ideal internal supply must be set to some value above the fixed value.
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