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?
-
A8 V
-
B320 V
-
C3.2 V
-
D32 V
Answer
Correct Answer: 32 V
Explanation
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:
Find R: R = V₁ / I₁ = 24 / 0.12 = 200 Ω.Compute new voltage: V₂ = I₂ * R = 0.16 * 200 = 32 V.Thus, set the source to 32 V.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