Answer:
R = 4Ω
Explanation:
If we have two resistors with resistances R1 and R2 in series the total resistance is R = R1 + R2
If the resistances are in parallel, the total resistance is given by:
1/R = 1/R1 + 1/R2.
First, we have a resistor with R1 = 1.5Ω
This resistor is connected in series with a parallel part (let's find the resistance of this parallel part), in one branch we have two resistors in series with resistances:
R2 = 8Ω and R3 = 4Ω
Because these are in series, the resistance of that branch is:
R = 8Ω + 4Ω = 12Ω
In the other branch, we have a single resistor of R4 = 4Ω
The resistance of the parallel part is:
1/R = 1/12Ω + 1/4Ω = 1/12Ω + 3/12Ω = 4/12Ω = 1/3Ω
1/R = 1/3Ω
R = 3Ω
Then we have a resistor (the first one, R1 = 1.5Ω) in series with a resistor of 3Ω.
Then the total resistance is:
R = 1Ω + 3Ω = 4Ω
