If the voltage across a fixed resistor is doubled and the current through that resistor also doubles, what can you conclude about the resistor's value?

Introduction:
This question checks proportional reasoning with Ohm's law. For a linear resistor, current is directly proportional to applied voltage, with resistance as the proportionality constant. If both voltage and current scale by the same factor, resistance remains the same.

Given Data / Assumptions:

• Linear, ohmic resistor behavior over the operating region.
• Initial and final operating points are within the resistor's power rating (no thermal runaway changing R).

Concept / Approach:
Ohm's law: R = V / I. If V is multiplied by k and I is multiplied by the same k, then R_new = (k * V) / (k * I) = V / I = R_old.

Step-by-Step Solution:
Start with R = V / I.Given: V' = 2 * V and I' = 2 * I.Compute: R' = V' / I' = (2 * V) / (2 * I) = V / I = R.Conclusion: Resistance value is unchanged.

Verification / Alternative check:
Example: A 1 kΩ resistor at 5 V draws 5 mA. Doubling V to 10 V yields 10 mA; R remains 1 kΩ. Measured data of typical metal-film resistors confirm near-constant R over moderate temperature rise.

Why Other Options Are Wrong:

• decreased/increased: Would require I to scale differently than V.
• impossible to determine: Direct calculation shows R is unchanged.
• became non-linear: Behavior described is perfectly linear.

Common Pitfalls:

• Forgetting that R is a ratio, not a fixed function of V alone.
• Confusing power changes (which do double: P = V * I) with resistance change.

Final Answer:
the resistor value did not change