A 2.0-ohm resistor is connected in series with a 20.0-V battery and a three-branch parallel network with branches whose resistances are 8.0 ohms each. Ignoring the battery's internal resistance, what is the current in the battery?

Answer 1

First, we can find the equivalent resistance of the parallel network. The reciprocal of the total resistance in a parallel network is equal to the sum of the reciprocals of the individual resistances:

1/R_total = 1/R1 + 1/R2 + 1/R3

Since all three branches have the same resistance (8.0 ohms), we can rewrite this as:

1/R_total = 3*(1/8)

1/R_total = 3/8

So, the reciprocal of the total resistance is 3/8, which means the total resistance in the parallel network is:

R_total = 8/3 ohms (approximately 2.67 ohms)

Now, we can combine this resistance in series with the 2.0-ohm resistor:

R_combined = R_total + R_series = (8/3) + 2 = (8+6)/3 = 14/3 ohms (approximately 4.67 ohms)

Finally, we can find the current in the battery using Ohm's Law:

I = V/R = 20.0 V / (14/3) ohms ≈ 4.29 A

So, the current in the battery is approximately 4.29 amps.