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

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

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

1. Compute the equivalent resistance of the network 2. Find the current in the 1.00 Ω resistor. 3. Find the current in the 3.00 Ω resistor. 4. Find the current in the 7.00 Ω resistor. 5. Find the current in the 5.00 Ω resistor.

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!

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