To isolate $R_{eq}$ in the equation

$\frac{1}{R_{eq}} = \frac{1}{r_1} + \frac{1}{r_2},$

follow these steps:

1. Start with the given equation: $\frac{1}{R_{eq}} = \frac{1}{r_1} + \frac{1}{r_2}$

2. Find a common denominator on the right-hand side: $\frac{1}{R_{eq}} = \frac{r_2 + r_1}{r_1 r_2}$

3. Take the reciprocal of both sides to solve for $R_{eq}$: $R_{eq} = \frac{r_1 r_2}{r_1 + r_2}$

So, the equivalent resistance $R_{eq}$ is given by

$R_{eq} = \frac{r_1 r_2}{r_1 + r_2}.$

Would you like further details or have any questions?

Here are 5 related questions to consider:

1. How does the equivalent resistance $R_{eq}$ change as $r_1$ and $r_2$ increase?
2. What is the physical meaning of $R_{eq}$ in a parallel circuit?
3. How does this formula change if $r_1$ and $r_2$ are identical?
4. How would you extend this formula to three resistors in parallel?
5. What would happen to $R_{eq}$ if one of the resistances $r_1$ or $r_2$ approaches infinity?

Tip: When resistors are connected in parallel, the equivalent resistance is always lower than the smallest individual resistor in the network.