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 $r_1$ and $r_2$ are connected in parallel. find the total resistance when $r_1 = x$ and $r_2 = x + 1$. $\frac{1}{2x + 1};x
eq - 1,-\frac{1}{2},0$ $2x + 1;x
eq - 1,0$ $\frac{x(x + 1)}{2x+1};x
eq - 1,-\frac{1}{2},0$ $\frac{2x + 1}{x(x + 1)};x
eq - 1,0$

Explanation:

Step1: Recall parallel - resistance formula

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

Step2: Find a common denominator

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

Step3: Solve for $r$

Taking the reciprocal of both sides, we get $r=\frac{x(x + 1)}{2x+1}$, where $x
eq - 1,0$ (because when $x=-1$, $r_2 = 0$ and when $x = 0$, $r_1=0$, and resistance cannot be zero in this context, and also to avoid division by zero in the original formula for parallel resistance).

Answer:

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