Series opposing sources: two 1.5 V cells are connected in series opposition across two 100 Ω resistors in series. What is the total current in the circuit?

Introduction / Context:
Battery sources can be connected in series aiding (voltages add) or series opposing (voltages subtract). Understanding net source voltage is essential before applying Ohm’s law to find current in a simple series network.

Given Data / Assumptions:

• Two cells: each 1.5 V.
• Connection: series opposing (opposite polarity).
• Load: two 100 Ω resistors in series (total 200 Ω).
• Ideal sources and resistors (no internal resistance considered).

Concept / Approach:

In series opposition, the net voltage equals the difference of the two cell voltages. Since both are 1.5 V, the net source becomes zero volts. With zero applied voltage across a finite resistance, circuit current is zero amperes.

Step-by-Step Solution:

Verification / Alternative check:

Kirchhoff’s voltage law perspective: opposing sources cancel, leaving no driving emf, hence no current in a closed loop of finite resistance.

Why Other Options Are Wrong:

Any nonzero current would require a nonzero net voltage or a short (zero resistance), neither is present per the data.

Common Pitfalls:

Accidentally adding the voltages as if they were aiding; forgetting that equal and opposite sources fully cancel.

Final Answer:

0 A