Source stacking: Two DC supplies of 5 V and 16 V are connected in series-opposing. What is the resulting total voltage seen by the load?

Introduction / Context:Combining voltage sources in series is common in test setups and battery packs. Whether sources are series-aiding or series-opposing determines the net voltage delivered to the circuit. Understanding the sign convention prevents accidental under- or over-voltage conditions.

Given Data / Assumptions:

• Ideal DC sources: 5 V and 16 V.
• Series-opposing orientation (polarities oppose).
• No internal resistance considered (idealization).

Concept / Approach:For series-aiding, voltages add. For series-opposing, the net is the difference between magnitudes, with the sign of the larger source. Thus total magnitude = |16 − 5| = 11 V, polarity following the 16 V source orientation.

Step-by-Step Solution:

Verification / Alternative check:Apply KVL around the loop, summing rises and drops with proper signs; the algebra yields a net of +11 V in the larger source polarity.

Why Other Options Are Wrong:

• 16 V or 21 V: 16 V would ignore the 5 V opposing source; 21 V is for series-aiding.
• 80 V: Not related; likely a distractor.

Common Pitfalls:Confusing aiding vs opposing and forgetting to consider polarity when stacking sources.

Final Answer:11 V