Problem 3.7-2

• a. We have two resistors in parallel, so the 1^st equivalent resistor is: 1/R_eq1 = 1/6 + 1/3 = 3/6 = 1/2 so R_eq1 = 2 Ω.

  Then we have two resistors in series, one of 4Ω and Req1 of 2Ω so we get R1 = 2 + 4 = 6Ω.

• b. Now we have R1 in parallel with two resistors so we get R_eq2 such that
  1/R_eq2 = 1/R1 + 1/6 + 1/12 = 5/12 so we have R_eq2 = 12/5 Ω

  Then we have R_eq2 in series with a 8Ω resistor, thus we have: R2 = 12/5 + 8 = 52/5 Ω

• c. For the currents we have I = current in the 6Ω resistor = 2 - i1.
  We can apply the Kirchoff’s law for the voltages. We have:
  U (voltage source) = v1 + v2 (in the 6Ω resistor)

  So 24 = i1*R2 + 6*(i1-2) so we get i1 = 36*5/82 = 2.19 A

  Then we can find the voltage v1 = i1*R2 = (36*5/82)*(52/5) = 36*26/41 = 22.8 V