A2.0-ohm resistor is connected in a series with a 20.0 -v battery and a three-branch parallel network with branches whose resistance are 8.0 ohms each. ignoring the battery’s internal resistance, what is the current in the battery? show your work.

4.3A.

The easiest way to solve this problem is find the equivalent resistance for parallel resistor 1/Req = 1/R1 + 1/R2 + 1/R3 in the three-branch parallel network with branches whose resistance are 8Ω.

1/Req = 1/8 Ω + 1/8 Ω + 1/8 Ω

1/Req = 3/8 Ω

Req = 8/3 Ω = 2.667Ω

Req = 2.7Ω

So, the equivalent circuit will be the 20.0V battery in series with a resistor  2.0Ω and the equivalent resistor 2.7Ω.

Using Ohm's Law to find the current provide by the 20.0V voltage source:

V = I*R > I = V/R

I = 20.0V/(2.0Ω + 2.7Ω)

I = 20.0V/4.7Ω

I = 4.3A

the answer is false

explanation:

the answer is a. at the equator