Electronic – Resistors in series and parallel conundrum

dcparallelresistanceseries

Three resistors of some resistance (ohms) are put in series and the measured total resistance is 24 ohm. Now, if the same three resistors are connected in parallel to each other and the total resistance measured for this combination is Rp, would the value of Rp ever reach 1.8 ohm or 4.4 ohm?
Also how would you set-up the equations to solve such a question?

I have tried to use the general resistor formulas to solve such an question:

r1 + r2 + r3 = 24 ohm

1/r1 + 1/r2 + 1/r3 = 1/Rp

But I cannot come up with a logical reason for this question. Would you be able to help me solve this conundrum?

Best Answer

Start by assuming 2 of the 3 resistors have the same value (\$R_1\$) and so the series equation is: -

$$24 = 2R_1+R_2$$ And therefore

$$R_1 = \dfrac{24-R_2}{2}$$

Next substitute 2 lots of R1 in the parallel resistor equation: -

$$\dfrac{2}{24-R_2} +\dfrac{2}{24-R_2} + \dfrac{1}{R_2} = \dfrac{1}{1.8}$$

$$=\dfrac{4}{24-R_2} + \dfrac{1}{R_2} = \dfrac{1}{1.8}$$

$$=\dfrac{24-R_2+4R_2}{24R_2 - R_2^2} = \dfrac{1}{1.8}$$

Then manipulate the algebra and solve the implied quadratic equation to find a value of R2 of 2.7205 ohms (1st solution). This implies R1 = 10.63975 ohms.

In series (R1 + R1 + R2) they equal 24 ohms and, in parallel they equal 1.8000 ohms.

The 2nd solution to the quadratic yields R2 = 15.8795 ohms and this means R1 is 4.06025 ohms. In series they are 24 ohms and in parallel they are also 1.8 ohms.