We are given the following information.
Resistor: R₁ = 400 Ω
Resistor: R₂ = 600 Ω
Resistor: R₃ = 500 Ω
We are asked to find the equivalent resistance of the given circuit.
Notice that the resistors R₁ and R₂ are in parallel and this parallel combination is in series with resistor R₃.
First, let us find the parallel resistance of R₁ and R₂.
[tex]R_p=\frac{R_1\times R_2}{R_1+R_2}=\frac{400\times600}{400+600}=\frac{240000}{1000}=240\;\Omega[/tex]
Finally, let us find the series resistance of Rp and R₃.
[tex]\begin{gathered} R_{eq}=R_p+R_3 \\ R_{eq}=240+500 \\ R_{eq}=740\;\Omega \end{gathered}[/tex]
Therefore, the equivalent resistance of the given circuit is 740 Ω.
