Circuit Loading

Overview

A voltage source may droop when we use it to power a component that has a low resistance compared to the internal resistance of the supply. To overcome this we always try to have resistive loads that are much larger than the internal resistance.

Calculations

To understand why this is that case consider the circuit below On the left hand side we consider a perfect circuit with no internal resistance. In this case it is obvious its always able to supply 20V across the load. But lets consider the circuit on the right with a internal resistance.

In this case the current that flows through the internal resistor is

$\frac{V}{R}=\frac{20}{1000+5}=0.0199A$

Thus the voltage drop across the load is

$V=0.019\cdot 1000=19.9V$

Which is not the required 20V. However if we used a $10\Omega$ resistor

$\frac{V}{R}=\frac{20}{10+5}=1.333A$

$V=1.333\cdot 10=13.3V$

which is very far from the desired voltage.