Using Ohm’s law, how many ohms of resistance are required to allow a current of 720 µA to flow when an applied voltage of 3.6 kV is present (assume linear, dc conditions)?

Introduction / Context:Calculating resistance from a specified current and voltage is a direct application of Ohm’s law. Accurate unit conversion (kV to V, µA to A) is crucial to avoid orders-of-magnitude errors that could lead to unsafe component choices.

Given Data / Assumptions:

• Voltage V = 3.6 kV = 3600 V.
• Current I = 720 µA = 720 * 10^-6 A = 0.00072 A.
• Ideal resistive behavior; temperature effects neglected.

Concept / Approach:Ohm’s law: R = V / I. Convert quantities to SI units first, then divide. Always cross-check reasonableness by computing the implied power to ensure the resistor wattage is practical.

Step-by-Step Solution:Convert units: V = 3600 V; I = 0.00072 A.Compute resistance: R = V / I = 3600 / 0.00072.Perform division: 3600 / 0.00072 = 5,000,000 Ω.Express neatly: R = 5 MΩ.Optional power check: P = V * I = 3600 * 0.00072 = 2.592 W → select an adequate wattage rating and high-voltage resistor construction.

Verification / Alternative check:Recompute using scientific notation: 3.6 * 10^3 / 7.2 * 10^-4 = (3.6 / 7.2) * 10^(3 + 4) = 0.5 * 10^7 = 5 * 10^6 Ω.

Why Other Options Are Wrong:200 nΩ and 5 kΩ are far too low; they would allow enormous currents.

200 kΩ is too low by a factor of 25.

50 MΩ is ten times too high, limiting current to 72 µA, not 720 µA.

Common Pitfalls:Forgetting to convert microamps to amps or kilovolts to volts, producing errors of 10^3–10^6.

Final Answer:5 MΩ