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?


simulate this circuit – Schematic created using CircuitLab

  • 3
    \$\begingroup\$ You can't find the voltage (I assume you mean voltage, not current) across the first resistor without taking the second resistor into account. You must use both resistors, which is in fact the voltage divider formula. \$\endgroup\$
    – Oldfart
    Apr 2, 2020 at 19:34
  • 3
    \$\begingroup\$ If you apply Ohm's law to both resistor and solve for the voltage across R2 you get the voltage divider formula. That's how it is derived in the first place. \$\endgroup\$
    – Hilmar
    Apr 2, 2020 at 19:44
  • \$\begingroup\$ Solve 2 equations instead of 1? \$\endgroup\$
    – user16324
    Apr 2, 2020 at 20:09
  • 1
    \$\begingroup\$ Both resistors work to limit the current - you must find the equivalent resistance of them (Easy for 2 in series, you literally add them up) and then divide voltage by this to find the current through them. In series the voltage at different points differ but current is the same - because its the same current you KNOW that both resistance work to limit the current. However once current is found you can multiply the current by any resistor resistance to find the voltage drop across it. \$\endgroup\$
    – QuickishFM
    Apr 3, 2020 at 12:41

3 Answers 3


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.


You can calculate it using Ohm's law by applying it twice: once to R1+R2 to find the current in the entire circuit, then again to find the voltage across R2.

But it's much faster if you know the divider formula, especially if you are trying to calculate for an unknown R1 and R2. Usually you are trying to calculate the resistances required to produce a voltage rather than the voltage produced by two resistances.

After you simplify all the equations down again, you just get the divider formula. So it's all the same.

In general, don't mistake someone using one approach to mean that no other approaches will work, especially if the other approaches are more fundamental. They probably just take more work because they are more general thus more versatile. Or sometimes, there's multiple methods of equal complexity and the author just had to choose one and did not feel like going through every possible approach.

If your brain is full and have to pick between remembering the divider formula or Ohm's law...pick Ohm's Law. Every time. It's more fundamental and can be applied in more situations and the divider equation comes from it.

  • \$\begingroup\$ Thank you for your answer. Now it makes sense. As to remembering formulas, I'm not a big fan of memorizing. I like to get an intuitive feeling of it and it will come to me by itself. \$\endgroup\$
    – Vio Ariton
    Apr 2, 2020 at 19:43
  • 1
    \$\begingroup\$ @VioAriton But voltage divider formula is worth memorize because we use it very very often. \$\endgroup\$
    – G36
    Apr 2, 2020 at 19:49
  • 1
    \$\begingroup\$ @G36 Well, it's one of the few that are simple enough you can actually get a feel for it in terms of ratios of resistances rather than outright memorizing it. Just like the parallel and series equivalents of resistors, capacitors, and inductors. It'll end up sticking in the OP's brain sooner or later. It is inevitable. There is no escape. \$\endgroup\$
    – DKNguyen
    Apr 3, 2020 at 16:15




\$E_2= 0.009\times1500=13.63 volts\$.

\$E_T= 1.37+13.63=15volts\$.

  • \$\begingroup\$ I think adding some explanations on what you did would be better \$\endgroup\$
    – NAND
    Apr 6, 2020 at 19:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.