Problem 3.7-15

We can first see that the resistor of 30 Ω is short-circuited because current can go “around” it.

On the right hand of the circuit, 8 Ω and 24 Ω are in parallel, so we have an equivalent resistor of 1/Re1 = 1/8 + 1/24 = 1/6 so Re1 = 6Ω.

This Re1 is in series with the 4Ω resistor, so we get Re2 = 10 Ω. This new resistor is in parallel with R, which gives a new resistor 1/Re3 = 1/10 + 1/R =  (R + 10)/10R

Thus Re3 = 10R/(10+R)

Then Re3 and the 12 Ω resistor are in parallel too, thus we have Re4 such that:

and this Re4 is in series with the 5Ω resistor, such that we finally have:

R _eq = Re4 + 5 = 9 Ω so Re4 = 4Ω

thus we have:

4*(120 + 12*R + 10*R) = 120*R

which is 480 + 88*R = 120*R so 32*R = 480 so R = 15 Ω