Question

A 2.0-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.

Answer 1

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