Hello I am currently a beginner in the world of electrical engineering and would like some help on voltage-divider circuits.

My question mainly concerns the "loading down effect". Can someone please explain what this effect is and what effect it has on a simple voltage divider circuit?

Please try to keep it as simple as possible without going into too much detail, I want to know the main idea behind it.



simulate this circuit – Schematic created using CircuitLab

Here's an example, see if this makes sense.

Consider the circuit on the left. Assume the meter VM1 is perfect- it has infinite input resistance. The voltage VM1 measures is \$ 10V \cdot \$ \$R2 \over (R1 + R2) \$ = 5V

Consider the circuit on the right. The meter VM2 has an input resistance of 1M represented by an ideal meter in parallel with 1M ohms. The equivalent resistance of R4 in parallel with R5 is R4||R5 = \$ R4\cdot R5 \over (R4 + R5) \$ = 90.909K ohms. The voltage VM2 reads is then \$ 10V \cdot \$ \$R4 || R5 \over (R3 + R4 || R5) \$ = 4.76V

The loading down of the 1M resistance reduced the reading by about 5%.

If you've been doing Thévenin equivalents, this should come as no surprise, since the unloaded divider could be replaced by a 5V source with 50K in series, and 1M is 20 times the source resistance.

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