A 12.0 V battery drives a 1.00 ohm, a 6.00 ohm and a 13.0 ohm resistor connected in series. What current flows in the circuit?

1. In series the resistances add: Rs = 1.00 ohm + 6.00 ohm + 13.0 ohm = 20.0 ohm. 2. Ohm's law for the whole loop gives I = V / Rs. 3. I = 12.0 V / 20.0 ohm = 0.600 A. 4. That one value of 0.600 A flows through all three resistors, since a series string offers a single path. 5. The figure 14.9 A is what the same battery would deliver if the three resistors were in parallel, so it is the trap for misreading the connection. _Source: OpenStax College Physics (CC BY 4.0), Ch 21 "Circuits and DC Instruments", section 21.1 Resistors in Series and Parallel_