Resistance prob: Solve 1/R = 1/R1 + 1/R2 + 1/R3 for R1

[foahchon]

The problem specifies to solve for \(\displaystyle R_1\), but I can't quite seem to get it:

\(\displaystyle \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\)

I can only seem to get as far as:

\(\displaystyle {R_1}{R_2}{R_3} = R({R_2}{R_3} + {R_1}{R_3} + {R_1}{R_2})\)

Can anyone give me a nudge in the right direction? Much appreciated.

 

[galactus]

One way is to solve for \(\displaystyle \frac{1}{R_{1}\) first, then take the reciprocal.

\(\displaystyle \L\\\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\)

\(\displaystyle \L\\\frac{1}{R_{1}}=\frac{1}{R}-\frac{1}{R_{2}}-\frac{1}{R_{3}}\)

Combine fractions:

\(\displaystyle \L\\\frac{R_{2}-R}{RR_{2}}-\frac{1}{R_{3}}=\frac{1}{R_{1}}\)

Combine fractions again:

\(\displaystyle \L\\\frac{R_{3}(R_{2}-R)-RR_{2}}{RR_{2}R_{3}}=\frac{1}{R_{1}}\)

Reciprocal:

\(\displaystyle \H\\R_{1}=\frac{RR_{2}R_{3}}{R_{3}(R_{2}-R)-RR_{2}}\)

If you want to continue with your method, go ahead and get R1 on one side of the equation.

\(\displaystyle \L\\R_{1}R_{2}R_{3}=RR_{2}R_{3}+RR_{1}R_{3}+RR_{1}R_{2}\)

\(\displaystyle \L\\R_{1}R_{2}R_{3}-RR_{1}R_{3}-RR_{1}R_{2}=RR_{2}R_{3}\)

Factor out R1:

\(\displaystyle \L\\R_{1}(R_{2}R_{3}-RR_{3}-RR_{2})=RR_{2}R_{3}\)

Divide through:

\(\displaystyle \L\\R_{1}=\frac{RR_{2}R_{3}}{R_{2}R_{3}-RR_{3}-RR_{2}}\)

Factor the denominator:

\(\displaystyle \H\\\frac{RR_{2}R_{3}}{R_{3}(R_{2}-R)-RR_{2}}\)

See, same thing as above using the other method.

 

[foahchon]

You know, I knew it was going to be something really simple. I just couldn't figure out how to factor out \(\displaystyle R_1\), guess I didn't have my head on straight.

But anyway, thanks very much.