Question

multiple choice 1 point the equation (r=\frac{1}{\frac{1}{r_1}+\frac{1}{r_2}}) represents the total resistance, (r), when two resistors whose resistances are (r_1 = z) and (r_2=z + 1) are connected in parallel. find the total resistance when (r_1 = z) and (r_2=z + 1). options: (\frac{1}{2z + 1};z
eq - 1,-\frac{1}{2},0), (2z + 1;z
eq - 1,0), (\frac{z(z + 1)}{2z+1};z
eq - 1,-\frac{1}{2},0), (\frac{z + 1}{z(2z + 1)};z
eq - 1,0)

Explanation:

Step1: Recall parallel - resistance formula

The formula for the total resistance $r$ of two resistors $r_1$ and $r_2$ in parallel is $\frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}$. Given $r_1 = z$ and $r_2=z + 1$, we substitute these values into the formula: $\frac{1}{r}=\frac{1}{z}+\frac{1}{z + 1}$.

Step2: Find a common denominator

The common denominator of the right - hand side is $z(z + 1)$. So, $\frac{1}{r}=\frac{z + 1+z}{z(z + 1)}=\frac{2z + 1}{z(z + 1)}$.

Step3: Solve for $r$

Taking the reciprocal of both sides, we get $r=\frac{z(z + 1)}{2z + 1}$, where $z
eq-\frac{1}{2},0$.

Answer:

$\frac{z(z + 1)}{2z + 1};z
eq-\frac{1}{2},0$