Series-aiding sources (DC addition): Two ideal voltage sources of 6 V and 9 V are connected in series aiding. What is the total source voltage supplied to the circuit?

Difficulty: Easy

Correct Answer: 15 V

Explanation:


Introduction / Context:
When DC sources are connected in series aiding, their voltages add algebraically because the same current flows through both and their polarities support each other. This is common in battery packs and laboratory supplies arranged in series to reach a higher voltage.


Given Data / Assumptions:

  • Two ideal sources: 6 V and 9 V.
  • Series aiding polarity (positive of one tied to negative of the other so voltages add).
  • Neglect internal resistance for simplicity.


Concept / Approach:
Series-aiding voltages add directly: V_total = V1 + V2. If the sources were opposing, the difference would apply instead. Here, aiding is explicit, so addition is appropriate.


Step-by-Step Solution:
Identify polarity as aiding.Compute V_total = 6 V + 9 V.V_total = 15 V.


Verification / Alternative check:
Draw a simple loop and apply KVL: the sum of source rises equals the net rise, confirming 15 V available to the load.


Why Other Options Are Wrong:
(b) 16 V would require 7 V and 9 V or an error. (c) 7.5 V is an average, not applicable. (d) 8 V is unrelated. (e) Not needed as an exact value is known.


Common Pitfalls:
Mixing aiding with opposing; forgetting that source internal resistances, if present, affect current but not the ideal sum of voltages.


Final Answer:
15 V

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