Two 6.0 ohm resistors are connected in parallel, and that combination is in series with a 3.0 ohm resistor across a 12 V battery of negligible internal resistance. What current flows in each 6.0 ohm resistor?

1. Deal with the parallel pair first. Two equal resistors in parallel give half the value of one of them. 2. So the parallel combination has an equivalent resistance of 3.0 ohms. 3. That 3.0 ohms is in series with the other 3.0 ohm resistor, so the total resistance is 6.0 ohms. 4. The current from the battery is 12 volts divided by 6.0 ohms, which is 2.0 amperes. 5. The whole 2.0 amperes passes through the single series resistor, so the potential difference across it is 2.0 amperes multiplied by 3.0 ohms, or 6.0 volts. 6. The remaining potential difference across the parallel section is 12 volts minus 6.0 volts, which is 6.0 volts. 7. Each branch feels that same 6.0 volts, so the current in each 6.0 ohm resistor is 6.0 volts divided by 6.0 ohms, which is 1.0 ampere. 8. The 2.0 ampere answer is the total current and traps anyone who forgets that the current splits, the 4.0 ampere answer ignores the series resistor completely, and the 0.5 ampere answer splits the branch current a second time. _Source: Siyavula Physical Sciences Grade 12 (Everything Science, CC BY 4.0), Chapter 10: Electric circuits_