Voltage division in series: A 45 V DC source is applied to a series network with total resistance 3300 Ω. If one of the resistors, R3, is 1200 Ω, what is the voltage drop across R3?

Introduction / Context:
Voltage division in series circuits is a fundamental technique for predicting how a source voltage distributes across individual elements. This problem directly applies the voltage divider rule to determine the drop across one known resistor when total resistance and source voltage are given.

Given Data / Assumptions:

• Total series resistance, R_total = 3300 Ω.
• Source voltage, V_s = 45 V DC.
• Target resistor, R3 = 1200 Ω (in series).
• Ideal resistors; no additional elements.

Concept / Approach:
For series circuits, current I is the same through all resistors: I = V_s / R_total. The voltage across any resistor R_x is V_x = I * R_x. Combining, V_x = V_s * (R_x / R_total). This is the standard voltage divider formula.

Step-by-Step Solution:
Compute the ratio: R3 / R_total = 1200 / 3300 = 0.363636…Apply divider: V_R3 = 45 * 0.363636…Multiply: 45 * 0.363636… = 16.3636… V.Round sensibly to two decimals: V_R3 ≈ 16.36 V.

Verification / Alternative check:
Current method: I = 45 / 3300 = 0.013636… A. Then V_R3 = I * 1200 = 0.013636… * 1200 = 16.3636… V. Both methods agree.

Why Other Options Are Wrong:
32.72 V: would correspond to a much larger fraction (about 0.727) not matching 1200/3300.10.90 V: too small; implies R3 is roughly 800 Ω of the total, which it is not.15.00 V: a rounded guess, not the precise divider result.

Common Pitfalls:
Forgetting that only series elements share the same current; mixing up series and parallel results leads to incorrect voltage predictions. Ensure total resistance includes all series elements when using the divider rule.

Final Answer:
16.36 V