Problem 3.7-11

The first thing to do here is to find the equivalent resistor.

The rightmost branch is equivalent to R1 = 8Ω, the middle one to R2 = 8Ω also (fig. a)

Those two branches are in parallel so the equivalent resistor is 1/Re = 1/R1 + 1R2

So 1/Re = 1/8 + 1/8 = 1/4 so Re = 4Ω (fig .b)

            Fig. a                                                              Fig. b

Now this 4Ω resistor is in series with the 20Ω one, this is thus R3 = 24Ω.

This R3 is in parallel with the 12Ω resistor, thus 1/R_{eq a-b}  = 1/12 + 1/24 = 3/24 = 1/8

Thus R_{eq a-b} = 8 Ω

Now for the current i, we have v_ab = 40 V from which we can deduce the current in the rightmost branch of fig. b which is

Ir = v_ab /R3 = 40/24 = 5/3 A

As the two resistors on the right on fig. a have the same value (8Ω), the currents going through them are the same.

Thus we have Ir = 2*i thus i = Ir/2 = 5/6 A