For a fixed resistor R, if the current through it is halved in steady DC operation, what must happen to the voltage across the resistor (assume R remains constant)?

Introduction:
This question probes proportional reasoning with Ohm's law when resistance is fixed. It is common in bias network design and troubleshooting where current changes must be related to voltage changes across a known resistor.

Given Data / Assumptions:

• DC steady state.
• Resistor value R is constant (no temperature drift considered).
• Current is reduced to half its original value.

Concept / Approach:
Ohm's law gives V = I * R. With R constant, voltage is directly proportional to current. Therefore, halving current halves the voltage across the resistor as well.

Step-by-Step Solution:
1) Original voltage: V1 = I1 * R.2) New current: I2 = I1 / 2.3) New voltage: V2 = I2 * R = (I1 / 2) * R = V1 / 2.4) Hence, the voltage across the resistor is halved.

Verification / Alternative check:
Numeric example: R = 100 Ω, I1 = 100 mA → V1 = 10 V. If I2 = 50 mA, V2 = 5 V, precisely half.

Why Other Options Are Wrong:
The resistance is halved: contradicts the assumption of fixed R.
The voltage doubles: opposite of direct proportionality under constant R.
None of the above: incorrect because the correct relation is explicit.
Power halved but voltage unchanged: power P = V * I would not behave this way if V were unchanged while I halves (that would halve power, but then V would not satisfy V = I * R).

Common Pitfalls:
Mistaking proportionality: with constant R, V tracks I. Do not confuse with cases where the source voltage is fixed and current changes because R changes.

Final Answer:
the voltage is halved