Series-aiding sources with series resistors: Two 6 V batteries are connected in series aiding and applied across two 1.2 kΩ resistors connected in series. What is the current through each resistor (steady-state DC)?

Introduction / Context:Series circuits are foundational in basic electrical engineering. When sources are connected series aiding, their voltages algebraically add. In a pure series path, the same current flows through every element. This question reinforces applying Ohm's law to a composite of series sources and series resistors, and it checks that you remember current is identical through each series component.

Given Data / Assumptions:

• Batteries: two sources of 6 V each, connected series aiding.
• Resistors: two resistors of 1.2 kΩ each, connected in series.
• Assume ideal components (no internal resistance), steady DC conditions.

Concept / Approach:Combine series sources by addition and combine series resistors by summation. Then apply Ohm's law I = V / R for the total loop. In a series circuit, that same current passes through each resistor; therefore the computed loop current is the current through each resistor as well.

Step-by-Step Solution:

Verification / Alternative check:Voltage drops: V1 = I * 1.2 kΩ = 5 mA * 1200 Ω = 6 V; V2 = 6 V. Sum is 12 V, matching the source addition, so KVL is satisfied.

Why Other Options Are Wrong:

• 10 mA: Would require R_total = 1.2 kΩ, not 2.4 kΩ.
• 0 A: Only true for an open circuit, which is not the case here.
• 2.5 mA: Implies an incorrect total resistance or voltage sum.

Common Pitfalls:

• Forgetting that series aiding sources add and series resistors sum.
• Applying the current to only one resistor rather than the entire loop.

Final Answer:5 mA