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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?

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  • \$\begingroup\$ @Atizs If R1=infinity, then the 3 resistors in series can't be 24R. \$\endgroup\$
    – HandyHowie
    Apr 3, 2019 at 11:34
  • \$\begingroup\$ No r1 + r2 + r3 must equal 24 ohm, so the combination of the three resistors must add up to 24 ohms . For example, 8 + 8 + 8 = 24 or 6 + 6 +12 = 24 \$\endgroup\$
    – Nilp Amin
    Apr 3, 2019 at 11:37
  • \$\begingroup\$ 2.2 || 10.9 || 10.9 = 1.83 ohms. 2.2+10.9+10.9 = 24 \$\endgroup\$
    – Andy aka
    Apr 3, 2019 at 11:40
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    \$\begingroup\$ You have 3 unknowns and two equations. Note what Andy did. \$\endgroup\$
    – Spehro Pefhany
    Apr 3, 2019 at 11:54
  • \$\begingroup\$ Ooops my above calc is wrong but not far off hitting the mark. 3 || 10.5 || 10.5 gives 1.875 ohms and 24 ohms in series. \$\endgroup\$
    – Andy aka
    Apr 3, 2019 at 11:59

1 Answer 1

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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.

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  • \$\begingroup\$ Perfect. Thankyou all for the help! \$\endgroup\$
    – Nilp Amin
    Apr 3, 2019 at 12:26
  • \$\begingroup\$ All you need to do now is substitute the 1.8 ohms with 4.4 ohms and see what that comes up with. \$\endgroup\$
    – Andy aka
    Apr 3, 2019 at 12:26

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