Non-ideal sources in series-opposing: net voltage can still be less than the largest source due to voltage drops

• A. True
• B. False
• C. Only in reactive loads
• D. Only at high temperatures

Answer: (A)

When non-ideal sources with internal resistances are placed in series and oppose each other, the current that flows creates voltage drops inside each source. The voltage you measure across the outer terminals isn’t just the difference of the open-circuit EMFs anymore; it’s reduced by the drops across the internal resistances. That means the net terminal voltage can end up smaller than the largest individual source’s EMF.

For example, if you have a 12 V source with 1 Ω of internal resistance opposing a 7 V source with 1 Ω internal resistance, the difference is 5 V. The current that flows is 5 V divided by the total internal resistance (2 Ω), so I = 2.5 A. The internal drops are I*r for each source, which sum to 5 V; as you traverse the chain from one outer terminal to the other, these drops reduce the terminal voltage, which can even drop below the 12 V source’s rating. This illustrates why the net voltage can be less than the largest source in series-opposing arrangements with non-ideal sources.