A coil with a self-inductance of 0.90 mH experiences a constant current buildup from zero to 15 A in 0.65 s.What is the induced emf during this interval?
A)8.78E-3 V
B)2.08E-2 V
C)8.78E+2 V
D)2.56E-2 V
E)2.31E-2 V

Question

Quizplus · Accepted Answer

The induced emf ($ε$) in a coil due to a change in current ($ΔI$) over a time interval ($Δt$) can be calculated using the formula $ε = L \frac{ΔI}{Δt}$, where $L$ is the self-inductance of the coil. Substituting the given values: $ε = 0.90 \times 10^{-3} \frac{15 - 0}{0.65}$ results in $ε = 2.08 \times 10^{-2}$ V.

A Coil with a Self-Inductance of 0