Thevenin and loading
Using Thevenin to solve WS0 problems.
Consider this Ohm's law problem:
- $V_\text{th}$ = output voltage with no load $\stackrel{?}{=}$
- Short 12V to ground: There are now 2 parallel paths to ground. $R\text{th}\stackrel{?}{=}$
$V=4V + 1\text{ mA}\cdot 670\ \Omega=$ 4.67 V
That complicated Thevenin circuit.
Rule of 10 and loading
Consider the voltage divider below coupled to a load across $R_2$.
Typically the load is specified in terms of a desired voltage and desired current.
If we design the voltage divider for the desired load, then in order for the load not to pull down the voltage "too much" (stay within about 10%) then a rule of thumb is that...
Rule of 10
The current $I_2$ through $R_2$ should be roughly 10 times the desired load current.
Consider this circuit. Will it be able to maintain the voltage desired? If not, what would better values of $R_1$ and $R_2$ be?
