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My question is very basic. I have a resistance circuit in which two resistances of 2 ohms are connected in parallel. I want to find the equivalent resistance. I know the equivalent resistanve will be

$$R_{eq} = \frac{R_1 \cdot R_2}{R_1 + R_2} = 1 \Omega$$

My doubts resides at the unit. My teacher says, the unit is ohm. But what we have done is a division. Since I perform division on same entity (in this case it is resistance), then it should become a unit-less quantity right ? But at the same time it is known as equivalent resistance. That means it is a resistance and hence it's unit should be ohm. This confuses me.

So my question is, what will be the unit if we divide same entity (say resistance) or it would be unitless ??

Note: Sorry for this basic doubts. But I really need an answer. If anyone can remove my doubt then that will be really a great help.

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The equation for two resistors in parallel is $$R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2}$$

The numerator is a product of resistances so its unit is Ohms squared (\$\Omega^2\$).

The denominator is a sum of resistances so its unit is Ohms (\$\Omega\$).

Dividing the numerator by the denominator gives you Ohms: $$\frac{[\Omega^2]}{[\Omega]} = [\Omega]$$

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  • \$\begingroup\$ what happens if it is r1/r2 ?? what is the unit ?? \$\endgroup\$ – Rajeev K Tomy Jul 17 '15 at 19:59
  • \$\begingroup\$ @RajeevKTomy Your equation was wrong. \$\endgroup\$ – Null Jul 17 '15 at 20:01
  • \$\begingroup\$ @RajeevKTomy If you're dividing resistances then it would be unitless. \$\endgroup\$ – Null Jul 17 '15 at 20:01
  • \$\begingroup\$ yes that was a mistake when typing the question. corrected it. \$\endgroup\$ – Rajeev K Tomy Jul 17 '15 at 20:02
  • \$\begingroup\$ this confuses me. When I ask a question like this to teacher "what is 2ohm/1ohm" then teacher said it is 2Ohm itself and not 2. units will preserve in the division. Is that right ?? \$\endgroup\$ – Rajeev K Tomy Jul 17 '15 at 20:05
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The correct equation for two parallel resistances is: \begin{gather} \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \end{gather} Which can be simplified using Algebra to: \begin{gather} R_{eq} = \frac{R_1 R_2}{R_1 + R_2} \end{gather}

The numerator has units of \$\Omega^2\$ (because you multiplied two resistances), then dividing by a resistance gives \$\Omega\$, the correct units for resistance.

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  • \$\begingroup\$ what happens if it is r1/r2 ? what is the unit ? \$\endgroup\$ – Rajeev K Tomy Jul 17 '15 at 20:00
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    \$\begingroup\$ @RajeevKTomy Then it would be dimensionless (i.e. a ratio). \$\endgroup\$ – Andrey Akhmetov Jul 17 '15 at 20:04
  • \$\begingroup\$ this confuses me. When I ask a question like this to teacher "what is 2ohm/1ohm" then teacher said it is 2Ohm itself and not 2. units will preserve in the division. Is that right ?? \$\endgroup\$ – Rajeev K Tomy Jul 17 '15 at 20:07
  • \$\begingroup\$ @RajeevKTomy: Q: What do you get if you divide 10 apples in groups of 5 apples? A: 2 groups (not 2 apples). \$\endgroup\$ – Curd Jul 17 '15 at 21:03
  • \$\begingroup\$ @Curd 2 Apples per Apple, or 3 dBApple (that's decibels relative to one apple). \$\endgroup\$ – Tom Carpenter Jul 18 '15 at 1:16
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The resulting unit is Ohm.

You have the formula up-side-down, but it just happens to give the right answer with 2 Ohm resistors.

The correct formula is \$R_{eq} = \dfrac{R_1 \cdot R_2}{R_1 + R_2}\$, which gives units of \$\dfrac{\Omega^2}{\Omega} = \Omega\$.

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In a situation where you are actually dividing one parameter by another that has the same unit, the result in unitless or "dimensionless (i.e. a ratio)" as commented by hexafraction. If your teacher answered otherwise, he/she is either wrong or misunderstood the question. That is the situation if you divide the resistance to common by the total resistance of a voltage divider. The result is a ratio that you can then multiply by the supply voltage to get the output voltage.

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