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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 network2. 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.
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Jakob Littel

Feb 20, 2025

Both branches of the parallel circuit have two resistors in series.R1 = 1 + 3 = 4 ΩR2 = 7+ 5 = 12 Ω1/Req = ¼ + 1/12 = ⅓Req = 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 ampsThe same amount of current flows through the 1 Ω and 3 Ω resistors.I = 46.5 ÷ 12 = 3.875 ampsTotal current = 11.625 + 3.875 = 15.5 ampsLet’s divide the voltage by the equivalent resistance.I = 46.5 ÷ 3 = 15.5 ampsThis proves that the answers are correct. 4 + 12 = 164/16 = ¼ This is the fraction of the current that flows through the 1 Ω and 3 Ω resistors.¼ * 15.5 = 3.875 amps¾ * 15.5 = 11.625 ampsThis is simple way to solve this type of problem. I hope this is helpful for you.... Show More
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J
Jakob Littel

Feb 20, 2025

The source sees (1+3)||(7+5) = 48/16 = 3Ω. .Thus the amps entering the ckt = 46.5/3 = 31/2 = 15.5AThe currents in the 1 and 3Ω are the same and = 15.5*12/16 = 11.625AThe currents in the 7 and 5Ω are the same and = 15.5*4/16 = 3.875ACheck:11.625*(1+3) = 3.875*(7+5)? Yes!... Show More
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