Answer:
Approximately [tex]111\; {\rm \Omega}[/tex].
Explanation:
It is given that [tex]R_{1} = 200\; {\Omega}[/tex] and [tex]R_{2} = 250\; {\Omega}[/tex] are connected in a circuit in parallel.
Assume that this circuit is powered with a direct current power supply of voltage [tex]V[/tex].
Since [tex]R_{1}[/tex] and [tex]R_{2}[/tex] are connected in parallel, the voltage across the two resistors would both be [tex]V[/tex]. Thus, the current going through the two resistors would be [tex](V / R_{1})[/tex] and [tex](V / R_{2})[/tex], respectively.
Also because the two resistors are connected in parallel, the total current in this circuit would be the sum of the current in each resistor: [tex]I = (V / R_{1}) + (V / R_{2})[/tex].
In other words, if the voltage across this circuit is [tex]V[/tex], the total current in this circuit would be [tex]I = (V / R_{1}) + (V / R_{2})[/tex]. The (equivalent) resistance [tex]R[/tex] of this circuit would be:
[tex]\begin{aligned} R &= \frac{V}{I} \\ &= \frac{V}{(V / R_{1}) + (V / R_{2})} \\ &= \frac{1}{(1/R_{1}) + (1 / R_{2})}\end{aligned}[/tex].
Given that [tex]R_{1} = 200\; {\Omega}[/tex] and [tex]R_{2} = 250\; {\Omega}[/tex]:
[tex]\begin{aligned} R &= \frac{1}{(1/R_{1}) + (1 / R_{2})} \\ &= \frac{1}{(1/(200\: {\rm \Omega})) + (1/(250\; {\rm \Omega}))} \\ &\approx 111\; {\rm \Omega}\end{aligned}[/tex].
