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?
Correct Answer: 11 V
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:
Identify opposing: one source is reversed relative to the other.Compute magnitude: 16 − 5 = 11 V.Assign polarity: direction of the larger (16 V) dominates the net.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