Three-tap divider with equal resistors: A 6 V battery feeds three 2.2 kΩ resistors in series to create two output taps. What are the two tap voltages measured from the negative end (assume no load)?

Introduction / Context:
Voltage dividers distribute a source voltage across series resistors in proportion to their resistances. With equal-value resistors and no load, the drops are equal, yielding equally spaced output taps. This is a staple concept for bias networks and reference ladders.

Given Data / Assumptions:

• Source: 6 V DC.
• Three equal resistors: each 2.2 kΩ in series.
• Outputs: the intermediate node after the first resistor and the node after the second resistor (both referenced to the negative end).
• No load (open-circuit taps), so divider ratios are ideal.

Concept / Approach:

Total resistance divides the source proportionally. With equal parts, each section drops V_total/3. Thus the first tap is one section above ground, the second tap is two sections above ground, and the third node is the full source voltage at the top of the ladder.

Step-by-Step Solution:

Verification / Alternative check:

KCL/KVL analysis yields the same since the current is uniform and no load draws additional current. Measured voltages stack linearly in equal increments.

Why Other Options Are Wrong:

2 V, 2 V ignores the cumulative second drop. 2 V, 6 V and 4 V, 6 V incorrectly include the full source at a tap rather than only at the top of the string.

Common Pitfalls:

Mixing up node numbering; assuming 6 V appears at an intermediate tap; forgetting that taps are cumulative sums of equal drops.

Final Answer:

2 V, 4 V