Assume that E = 46.5 V. The battery has negligible internal resistance.

2 Answers

• Both branches of the parallel circuit have two resistors in series.

  R1 = 1 + 3 = 4 Ω

  R2 = 7+ 5 = 12 Ω

  1/Req = ¼ + 1/12 = ⅓

  Req = 3 Ω

  The voltage for each branches of the parallel is 46.5 volts. The same amount of current flows through the 1 Ω and 3 Ω resistors.

  I = 46.5 ÷ 4 = 11.625 amps

  The same amount of current flows through the 1 Ω and 3 Ω resistors.

  I = 46.5 ÷ 12 = 3.875 amps

  Total current = 11.625 + 3.875 = 15.5 amps

  Let’s divide the voltage by the equivalent resistance.

  I = 46.5 ÷ 3 = 15.5 amps

  This proves that the answers are correct.

  4 + 12 = 16

  4/16 = ¼

  This is the fraction of the current that flows through the 1 Ω and 3 Ω resistors.

  ¼ * 15.5 = 3.875 amps

  ¾ * 15.5 = 11.625 amps

  This is simple way to solve this type of problem. I hope this is helpful for you.

• The source sees (1+3)||(7+5) = 48/16 = 3Ω. .

  Thus the amps entering the ckt = 46.5/3 = 31/2 = 15.5A

  The currents in the 1 and 3Ω are the same and = 15.5*12/16 = 11.625A

  The currents in the 7 and 5Ω are the same and = 15.5*4/16 = 3.875A

  Check:

  11.625*(1+3) = 3.875*(7+5)? Yes!