# 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.

• 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!