Series–parallel resistance between two nodes A and B A network consists of two ideal resistors connected in simple series between nodes A and B: a 10 Ω resistor followed by a 1 kΩ (1000 Ω) resistor. For this explicit series connection, what is the total equivalent resistance RT seen between A and B?

Introduction / Context:
This question checks your ability to compute the equivalent resistance of a strictly series connection. Series networks are everywhere in bias chains, current-limiting paths, and sensor dividers. Getting comfortable with series addition is essential before moving to mixed series–parallel reductions and Thevenin/Norton modeling.

Given Data / Assumptions:

• Two ideal resistors are connected in series between nodes A and B.
• R1 = 10 Ω and R2 = 1 kΩ (1000 Ω).
• Conductors are ideal; there are no hidden branches or parallel paths.
• We seek RT as seen between A and B.

Concept / Approach:
For resistors in series, the equivalent resistance is the arithmetic sum of the individual resistances. The rule is topology-based and independent of excitation (DC or AC) as long as elements are linear resistors. The governing relation is: R_total = R1 + R2 + … + Rn. Once RT is known, any loop current under a given source follows directly from I = V / RT.

Step-by-Step Solution:

Verification / Alternative check:
Use a quick current check. With a 10 V ideal source across A–B, I = 10 / 1010 ≈ 9.90 mA. The drops would be V_R1 ≈ 0.099 V and V_R2 ≈ 9.90 V, which sum back to 10 V, confirming series behavior and the calculated RT.

Why Other Options Are Wrong:

Common Pitfalls:
Confusing series with parallel (parallel uses reciprocals); mixing kΩ and Ω without converting; accidentally omitting a component or misreading comma formatting in values.

Final Answer:
1,010 Ω