Apply Ohm’s law: when 12 mA flows through a 1.2 kΩ resistor, determine the voltage drop across the resistor (express in volts).

Introduction / Context:
Voltage across a resistor is the product of current and resistance. This direct application of Ohm’s law appears constantly in biasing networks and sensor readouts, where small currents through kilo-ohm resistors create measurable voltages.

Given Data / Assumptions:

• I = 12 mA = 0.012 A.
• R = 1.2 kΩ = 1,200 Ω.
• Use V = I * R.

Concept / Approach:

Convert units carefully to maintain consistency, then multiply. Since both values are exact in this context, no rounding beyond standard decimal arithmetic is required.

Step-by-Step Solution:

Verification / Alternative check:

Using prefixes directly: 12 mA * 1.2 kΩ = (12 × 10^-3) * (1.2 × 10^3) = 14.4 × 10^0 = 14.4 V.

Why Other Options Are Wrong:

1.4 V and 10 V do not satisfy V = I * R with the given values; 100 V is far too large for milliampere currents through kilo-ohm resistances.

Common Pitfalls:

Mixing milli and kilo prefixes; dropping a power of ten; forgetting that volts scale linearly with both I and R.

Final Answer:

14.4 V