I'm currently working my way through Electronics for Dummies, where I am shown a simple voltage divider circuit consisting of a 9V battery and two resistors, R1 (12,000 ohm) and R2 (15,000 ohm).

Given the voltage divider as shown in the schematic below, and the voltage divider equation to calculate the output voltage (Voltage out) given as:

\$V_\text{out} = \left[ \frac{15000}{12000~+~15000} \right] \cdot 9V\$


simulate this circuit – Schematic created using CircuitLab


Why is the equation (shown above) using 15,000 as the numerator? Seeing as how the voltage output is after R1 and before R2, shouldn't we be using 12,000 as the numerator in the equation to calculate Vout?

Why then Does the equation above use \$\left[ \frac{15000}{12000~+~15000} \right] \cdot 9V\$?

  • \$\begingroup\$ You should always name a reference node i.e. name the -ve of the battery as 0V. This prevents doubts and reminds always that an arbitrary measurement of 4.5678 volts is relative to 0V. \$\endgroup\$
    – Andy aka
    Jun 25, 2014 at 11:04
  • \$\begingroup\$ Noted, i've changed the schematic to include 0V. \$\endgroup\$
    – Kenneth .J
    Jun 25, 2014 at 11:06

1 Answer 1


The numerator is the 15K resistor because you are measuring the voltage across R2 -- it is assumed the positive probe of the voltmeter is at "Voltage Out", and the negative probe of the meter (not shown) is at the bottom of R2 (the - side of the 9V battery).

The current through R2 is set up by ohm's law:

\$I = \frac{V}{R}\$, so \$I = \frac{9v}{12K + 15K}\$ = 333 µA

So the voltage across R2, using \$V = IR, V = 333 µA * 15K = 5v\$

Combining the two equations, you get \$V = \frac{9v}{12K + 15K} * 15K\$

which is the same as the equation in the last line of your question: \$V = \frac{15K}{12K + 15K}*9v\$

  • \$\begingroup\$ I see, i thought that the output voltage(Voltage Out) meant that they were measuring the voltage at that point, and not across R2(voltage drop). I guess i simply mis-read the book then. Thanks for clarifying my doubts. \$\endgroup\$
    – Kenneth .J
    Jun 25, 2014 at 11:10
  • \$\begingroup\$ The voltage at 'Voltage Out' (referenced to the 0v battery terminal) is the same as the voltage across R2. There isn't such a thing as 'voltage at' - it is alway 'voltage over' (between). If not specific, it generally means voltage between measure point and 0v. \$\endgroup\$
    – RJR
    Jun 25, 2014 at 12:17

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