Question

18. 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 batter? Show your work.

Answer 1

Here is the answer.  First get the impedance of the 3 parallel 8 ohm resistors, then add in the series resistor and solve with Ohm's Law

Answer 2

the correct answer would be 4.28

8ll8ll8 =1/ 1/8 /1/8 /1/8 = 2.67

2+2.67=4.67

20v/4.67=4.28A