Inductor voltage law: For a coil with L = 8 µH, if di/dt = 20 mA/µs, what is the induced voltage across the coil (v = L * di/dt)?
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A1.6 mV
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B160 mV
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C16 mV
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D2.5 mV
Answer
Correct Answer: 160 mV
Explanation
Introduction / Context:The defining inductor relation v = L * di/dt directly links voltage to how quickly current changes. Mastering unit conversions (µH, mA/µs) is critical for translating lab measurements into correct voltage estimates, especially in switching converters and transient analysis.
Given Data / Assumptions:
- L = 8 µH = 8 × 10^-6 H.
- di/dt = 20 mA/µs = 20 × 10^-3 A per 10^-6 s = 2.0 × 10^4 A/s.
- Ideal inductor; polarity per Lenz's law is not required for magnitude.
Concept / Approach:Apply the formula v = L * di/dt using consistent SI units. Convert microhenry to henry and milliampere-per-microsecond to ampere-per-second before multiplying to avoid powers-of-ten errors.
Step-by-Step Solution:
Convert L: 8 µH = 8e-6 H.Convert di/dt: 20 mA/µs = 0.02 A / 1e-6 s = 2.0e4 A/s.Compute v: v = 8e-6 * 2.0e4 = 1.6e-1 V = 0.16 V = 160 mV.Verification / Alternative check:Scale check: 10 µH * 10 mA/µs → 0.1 V. Our L and di/dt are close, giving ≈0.16 V, which is consistent by proportionality.
Why Other Options Are Wrong:
- 16 mV or 1.6 mV or 2.5 mV: These miss factors of 10^1–10^2 due to unit-conversion mistakes.
Common Pitfalls:
- Treating mA/µs as mA/s or forgetting to convert µH to H.
Final Answer:160 mV