Answer:
Rt = 908.25 [ohm]
Explanation:
In order to solve this problem, we must remember that the resistors connected in series are added up arithmetically.
In this case, R2 and R3 are in series therefore.
R₂₃ = 200 + 470
R₂₃ = 670 [ohm]
Now this new resistor (R₂₃) is connected in parallel with the resistor R4. therefore we must use the following arithmetic expression, to add resistances in parallel.
[tex]\frac{1}{R_{4-23} }= \frac{1}{R_{4}}+\frac{1}{R_{23} } \\\frac{1}{R_{4-23} }=\frac{1}{1800}+\frac{1}{670} \\R_{4-23}=488.25[ohm][/tex]
In this way R₁, R₅ and R₄₋₂₃ are connected in series.
Rt = R₁ + R₅ + R₄₋₂₃
Rt = 150 + 270 + 488.25
Rt = 908.25 [ohm]
