Physics

Physics

Problem

A circular loop contains two identical 9 V9\,\text{V} batteries and a resistor. Initially both batteries are oriented with their positive terminals pointing clockwise and the loop is closed. One battery is removed and re-inserted with its orientation reversed. Determine the direction—clockwise or counter-clockwise—of the conventional current in the loop after the reversal and justify your answer.
Intuitive explanation

The key insight is that reversing one battery flips its contribution to the net electromotive force (EMF) around the loop. Initially, both batteries add their voltages in the same direction, creating a net EMF of 18 V18\,\text{V} clockwise. When you flip one battery, it now opposes the other instead of aiding it—so the net EMF becomes 9−9=0 V9 - 9 = 0\,\text{V}. With zero net EMF, there's no driving force for current, so the conventional current stops flowing entirely.

Intuitive explanation

The key insight is that reversing one battery flips its contribution to the net electromotive force (EMF) around the loop. Initially, both batteries add their voltages in the same direction, creating a net EMF of 18 V18\,\text{V} clockwise. When you flip one battery, it now opposes the other instead of aiding it—so the net EMF becomes 9−9=0 V9 - 9 = 0\,\text{V}. With zero net EMF, there's no driving force for current, so the conventional current stops flowing entirely.

Browse the problem library