Voltage division in series: a 5.6 kΩ resistor and a 4.7 kΩ resistor are in series. If the voltage across the 5.6 kΩ resistor is 10 V, what is the voltage across the 4.7 kΩ resistor (assume the same series current)?

Introduction / Context:In a series network, the same current flows through all resistors. Voltage drops are proportional to resistance values, a principle known as the voltage divider. If one drop is known, the other can be found using ratios without knowing the total source voltage explicitly.

Given Data / Assumptions:

• Series resistors: R₁ = 5.6 kΩ, R₂ = 4.7 kΩ.
• Measured V₁ across 5.6 kΩ is 10 V.
• Ideal components and steady-state DC.

Concept / Approach:

In series: I is common. Therefore V ∝ R. The voltage across the second resistor is V₂ = V₁ * (R₂ / R₁). No need to know the source voltage; the ratio suffices.

Step-by-Step Solution:

Verification / Alternative check:

If I is common, V₁/V₂ = R₁/R₂ = 5.6/4.7 ≈ 1.1915; with V₁ = 10 V, V₂ ≈ 10 / 1.1915 ≈ 8.39 V, confirming the result.

Why Other Options Are Wrong:

10 V would imply equal resistances. 2.32 V is far too small for nearly equal-kilo-ohm values. 0 V is impossible with nonzero resistance and nonzero current.

Common Pitfalls:

Using R₁ + R₂ incorrectly; forgetting that the series current is the same and that voltages divide in proportion to resistances.

Final Answer:

8.39 V