Series resistance addition (basic recall): If a 47 Ω resistor is connected in series with a 68 Ω resistor, what is the total equivalent resistance of the series combination?

Introduction / Context:
In a series circuit, resistances add directly because the same current must pass through each element. This question checks your ability to compute an equivalent resistance quickly and to distinguish series behavior from parallel behavior, where the rule is different.

Given Data / Assumptions:

• Two resistors: 47 Ω and 68 Ω.
• They are connected in series across some source.
• Ideal components, DC conditions.

Concept / Approach:
For series elements, use the formula R_total = R1 + R2 + ... because the same current flows and voltage drops add. There is no current splitting in series paths, unlike in parallel networks.

Step-by-Step Solution:

Verification / Alternative check:
If these were in parallel, the equivalent would be less than the smallest resistor. Since our answer is greater than either alone, that is consistent with series behavior.

Why Other Options Are Wrong:
3,196 Ω is the product (47*68) and not applicable in series. “Less than 47 Ω/68 Ω” would describe a parallel equivalent, not series. 1,115 Ω is a miscalculation by adding an extra digit.

Common Pitfalls:
Mixing the series and parallel formulas, or mistakenly using the product-over-sum rule (valid only for two resistors in parallel).

Final Answer:
115 Ω