Parallel resistanee is reciprocal of the sum?

[Lokhtar]

Homework Statement

Why is parallel resistance the reciprocal of all individual resistances?

Homework Equations

V=IR

The Attempt at a Solution

Well, since V is constant and I is different, you can write it as I=V/R, and since V won't change, you can make it I=V*(1/R1+1/R2),etc. So I get that, but why do you then have to take the reciprocal of all the resistances to get the total resistance? Wouldn't it just be the direct sum of the individual 1/Rs??

 

[nrqed]

Homework Statement

Why is parallel resistance the reciprocal of all individual resistances?

Homework Equations

V=IR

The Attempt at a Solution

Well, since V is constant and I is different, you can write it as I=V/R, and since V won't change, you can make it I=V*(1/R1+1/R2),etc. So I get that, but why do you then have to take the reciprocal of all the resistances to get the total resistance? Wouldn't it just be the direct sum of the individual 1/Rs??

Because, by definition, $I = V/R_{eq}$. So setting this equal to your expression, we get

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

 

[nlightner]

The reason why parallel resistance is the reciprocal of all individual resistances is because of the way current flows in parallel circuits. In a parallel circuit, the current is divided among the different branches, and the total current is equal to the sum of the currents in each branch. This means that the total current is inversely proportional to the total resistance, as seen in the equation I=V/R. So, in order to calculate the total resistance, we need to take the reciprocal of all the individual resistances, which represents the inverse relationship between current and resistance in parallel circuits. In other words, the more resistors you add in parallel, the lower the total resistance will be, since the current is divided among more branches. This is why parallel resistance is the reciprocal of all individual resistances.