Okay in the image there is one battery (3.0V) facing the positive terminal up, then a resistor of 18-ohm, and another battery (6.0V) with the positive terminal also facing up.
I added the voltages (9.0V) and divided it by the resistor (18 ohm) to get a current of 0.5 A
The answer, however divided (3.0V) by (18 ohm) to get .17 A
What did they do?
Did they just use one battery?
Did they subtract the voltage?
I added the voltages (9.0V) and divided it by the resistor (18 ohm) to get a current of 0.5 A
The answer, however divided (3.0V) by (18 ohm) to get .17 A
What did they do?
Did they just use one battery?
Did they subtract the voltage?
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The two batteries oppose each other, so the net voltage across the resister is 3 volts.
So the current is (6v -3v) / 18 = 0.166A
So the current is (6v -3v) / 18 = 0.166A