Series–parallel resistive network: Two 3.3 kΩ resistors are connected in series, and this series combination is placed in parallel with a 4.7 kΩ resistor. If the voltage across one 3.3 kΩ resistor is 12 V, what is the voltage across the 4.7 kΩ resistor (steady DC conditions)?

Introduction / Context:
This problem checks understanding of how voltage behaves in series–parallel resistor networks. In DC circuits, series elements share the same current and their voltages add, while parallel elements share the same voltage across the branch terminals. Correctly identifying which components are series and which are parallel is the key to determining unknown voltages.

Given Data / Assumptions:

• A series string of two 3.3 kΩ resistors.
• This series string is in parallel with a single 4.7 kΩ resistor.
• The measured voltage across one 3.3 kΩ resistor is 12 V.
• Ideal resistors, steady-state direct current (no internal source resistance considered).

Concept / Approach:
In series, voltages add: V_series = V_R1 + V_R2. In parallel, both branches share the same terminal-to-terminal voltage. Therefore, the total voltage across the series pair equals the voltage across the parallel 4.7 kΩ resistor.

Step-by-Step Solution:

Verification / Alternative check:
Ohm’s law on the series branch shows equal-resistance, equal-drop behavior: if current is I, then V_each = I * 3.3 kΩ. Identical resistors in series share equal drops; hence two such drops total 24 V, the same voltage appearing across any parallel branch.

Why Other Options Are Wrong:

• 12 V or 6 V: These ignore that the 4.7 kΩ is across the entire two-resistor series branch, not across a single 3.3 kΩ element.
• 0 V: Would imply a short circuit across the branch, which is not stated.

Common Pitfalls:

• Mistaking the 4.7 kΩ as being in series with a single 3.3 kΩ rather than in parallel with the pair.
• Not adding series drops before assigning the parallel voltage.

Final Answer:
24 V