Electronic – Why is the voltage divider formula used in this circuit to find the terminal voltage

resistorsvoltagevoltage-source

I'm going through the book Practical Electronics for Inventors and there is a chapter about voltage sources that talks about practical voltage sources.

It calculates the terminal voltage of the circuit below using the voltage divider formula.

Why is the voltage divider formula used since the resistors are in series and why can't we calculate it using ohm's law, i.e find the current across the first resistor and multiply it with the second resistor?

schematic

simulate this circuit – Schematic created using CircuitLab

Best Answer

You are right that we can use Ohm's law.

Look at the math:

The current in this series circuit is:

$$I = \frac{V_1}{R_1+R_2} $$

And the voltage drop across \$R_2\$ is equal to:

$$V_{R_2} = I \times R_2 = \frac{V_1}{R_1+R_2} \times R_2 = V_1 \times \frac{R_2}{R_1 + R_2}$$

$$V_{R_1} = I \times R_1 = \frac{V_1}{R_1+R_2} \times R_1 = V_1 \times \frac{R_1}{R_1 + R_2}$$

As you can see the voltage divider formula comes directly from Ohm's law.