I'm currently working my way through electronics for dummies, where I am shown a simple voltage divider circuit comprising of a 6V battery and 2 resistors, R1 (220 ohm) and R2 (1000 ohm) in series.

To calculate the current, I will be using I = V (battery) / (R1 + R2)

As V1 can be calculated by the equation V1 = I x R1, by subbing this into the equation above, we get:

Equation 1

V1 = [ V(battery) / (R1 + R2) ] x R1

This is where I begin having problems, as the book then rearranges the equation by saying that I can rearrange the terms, without changing the equation to get:

Equation 2

V1 = [R1 / (R1 + R2)] x V(battery)

The problem I have is that I have no idea how the conversion from Equation 1 to Equation 2 takes place.

I have a poor background in mathematics so that accounts for my inability to decipher this seemingly simple conversion.

Could anyone help me out by listing the steps in order, for the conversion?

P.S I'm not sure if this should be under mathematics, but I placed it here since it concerns Ohm's law.

  • 4
    \$\begingroup\$ This question appears to be off-topic because it is about mathematics and not electrical engineering. \$\endgroup\$
    – Andy aka
    Commented Jun 25, 2014 at 11:01
  • \$\begingroup\$ I disagree, since the question is about how to determine the value of a voltage in a voltage divider, and the maths required to find an answer is integral to the electrical engineering effort and is not a stand-alone mathematical problem. \$\endgroup\$
    – EM Fields
    Commented Jun 25, 2014 at 11:06
  • \$\begingroup\$ That's a standalone math problem because op can't figure out why \$\frac{a}{b}c=a\frac{c}{b}\$ \$\endgroup\$ Commented Jun 25, 2014 at 11:39
  • \$\begingroup\$ If he'd merely asked the question: "Does (a/b)c = a(c/b)?", then I'd agree. However, he did not, he asked a question about electrical engineering, of which which math is a part. BTW, how'd you manage to format that equation like that? I didn't think it was possible in a comment. \$\endgroup\$
    – EM Fields
    Commented Jun 25, 2014 at 22:07

2 Answers 2


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

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


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.

  • \$\begingroup\$ Thanks for replying, i assume that in the case of the equation above, it would be V(battery) x [1/(R1 + R2)] x R1, it would be the equivalent of your ( a x c x b )= ( b x c x a )example above, therefore i can simply switch it around then? \$\endgroup\$
    – Kenneth .J
    Commented Jun 25, 2014 at 11:10
  • \$\begingroup\$ Yes, that's right. \$\endgroup\$ Commented Jun 25, 2014 at 11:10

This is a much less confusing procedure to follow:

enter image description here


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