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 r
esistance are 8.0 ohms each. Ignoring the battery’s internal resistance, what is the current in the batter? Show your work.
2 answers:
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
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
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