Four resistors are connected in series across a 50 V source. The measured series current is 100 µA. If R1 = 12 kΩ, R2 = 47 kΩ, and R3 = 57 kΩ, what value of R4 is required to account for the observed current?

Introduction / Context:This problem applies Ohm's law and series-resistance addition to determine an unknown resistor in a series string given the total current and supply voltage. Such calculations are common in sensor ladders, bias networks, and precision dividers where one element value must be chosen to meet a target current.

Given Data / Assumptions:

• Series supply voltage Vs = 50 V.
• Series current I_total = 100 µA = 100 * 10^-6 A.
• Known resistors: R1 = 12 kΩ, R2 = 47 kΩ, R3 = 57 kΩ.
• Find R4 such that the current matches the given value.

Concept / Approach:In series, total resistance is the sum of individual resistances. Ohm's law gives R_total = Vs / I_total. The unknown is R4 = R_total - (R1 + R2 + R3). Careful unit conversion (kΩ to Ω, µA to A) avoids arithmetic errors.

Step-by-Step Solution:Compute total resistance: R_total = Vs / I_total = 50 V / (100 * 10^-6 A).R_total = 50 / 0.0001 = 500,000 Ω = 500 kΩ.Sum known resistors: R_known = 12 kΩ + 47 kΩ + 57 kΩ = 116 kΩ.Find the unknown: R4 = R_total - R_known = 500 kΩ - 116 kΩ = 384 kΩ.Therefore, R4 = 384 kΩ.

Verification / Alternative check:Check current with the found value: R_sum = 116 kΩ + 384 kΩ = 500 kΩ. Then I = Vs / R_sum = 50 V / 500 kΩ = 0.0001 A = 100 µA, which matches the given current. The result is self-consistent.

Why Other Options Are Wrong:

• 38.4 kΩ or 3.84 kΩ: Off by factors of 10 or 100; would yield much higher current than specified.
• 3.84 MΩ: Too large; would reduce current far below 100 µA.

Common Pitfalls:

• Mistaking microamps for milliamps, leading to a 1000× error.
• Forgetting that series resistances add linearly, not in reciprocals (which applies to parallel).

Final Answer:384 kΩ