Electronic – Voltage divider, how to calculate Vout

Multiplication of scalars is commutative and associative, in other words:

a \$\times\$ b = b \$\times\$ a

and

a \$\times\$ c \$\times\$ b = b \$\times\$ c \$\times\$ a.

So if a = V(battery) and b = R1 and c = 1/(R1 + R2) then you can swap a and b.

Try this website for algebra help.

What do I take into consideration when designing the circuit?

Think it through.

• The load voltage will be highest when the current is lowest
• The load voltage will be lowest when the current is highest.
• You're given the high and low load current specification as well as the high and low load voltage specification.

So, you have two points on the load line. This will allow you calculate the equivalent resistance (Thevenin resistance) of the voltage divider as well as the voltage division factor \$\frac{R_2}{R_1 + R_2}\$.

Can you take it from here?

I edited the question to remove the current fluctuation part. I was just warned to keep it in mind, it wasn't part of the actual question.

Since there are two resistor values to pick, you need two independent specifications. Certainly, one of those specifications is that the voltage must be within 20% of 6V when the load current is 500mA.

If you form the Thevenin equivalent circuit of the 15V source and voltage divider resistors, the voltage across the load resistor is seen to be given by

$$15V \frac{R_2}{R_1 + R_2} - I_L\cdot R_1||R2 $$

So, you need another specification. One approach would be to limit the open-circuit voltage to be 20% higher than the nominal voltage.

$$15V \frac{R_2}{R_1 + R_2} = 7.2V$$

Now you have two equations and two unknowns:

$$\frac{R_2}{R_1 + R_2} = \frac{7.2}{15} $$

$$R_1||R_2 = \frac{7.2 - 6}{0.5A}$$

This is just an example of choosing an additional reasonable constraint in order to uniquely specify the resistor values. You might reasonably choose a different additional constraint.