Adjusting a DC source to change current: a resistor draws 120 mA from a 24 V supply. If you want 160 mA through the same resistor, what new source voltage should you set?

Introduction / Context:
When only the source voltage is adjustable, current changes proportionally for a fixed resistance. This is a direct application of Ohm’s law, often used in lab supplies when targeting specific currents into a known load resistance.

Given Data / Assumptions:

• I₁ = 120 mA at V₁ = 24 V.
• Same resistor (R is constant).
• Desired current I₂ = 160 mA.

Concept / Approach:

First compute the resistance using the initial operating point. Then compute the required voltage for the new current using V = I * R. Because R is constant, V scales with I linearly.

Step-by-Step Solution:

Verification / Alternative check:

Proportionality: V₂ / V₁ = I₂ / I₁ = 160/120 = 4/3 → V₂ = (4/3) * 24 = 32 V. Same result confirms the calculation.

Why Other Options Are Wrong:

8 V and 3.2 V would reduce current dramatically; 320 V is 10× too high and unsafe for typical bench circuits.

Common Pitfalls:

Forgetting to hold resistance constant; mixing mA and A; arithmetic errors in proportional scaling.

Final Answer:

32 V