Question

This is also for a Digital Electronics class​

Answer 1

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.

$\frac{1}{R_{4-23} }= \frac{1}{R_{4}}+\frac{1}{R_{23} } \\\frac{1}{R_{4-23} }=\frac{1}{1800}+\frac{1}{670} \\R_{4-23}=488.25[ohm]$

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]