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Inductor Transient Voltage (Electrical Engineering) Solution

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Problem Overview

We are given an inductor with inductance L=0.5 HL = 0.5\text{ H} and a rational current profile i(t)=6t Ai(t) = \frac{6}{t}\text{ A}. We evaluate didt\frac{di}{dt} from first principles and calculate the induced voltage VL(t)=LdidtV_L(t) = L \frac{di}{dt}.

Step-by-Step Mathematical Solution

Step 1: Set up the Difference Quotient Limit for didt\frac{di}{dt}

didt=lim⁡h→0i(t+h)−i(t)h=lim⁡h→06t+h−6th\frac{di}{dt} = \lim_{h \to 0} \frac{i(t+h) - i(t)}{h} = \lim_{h \to 0} \frac{\frac{6}{t+h} - \frac{6}{t}}{h}

Step 2: Combine Numerator Fractions using Common Denominator t(t+h)t(t+h)

didt=lim⁡h→06t−6(t+h)t(t+h)h\frac{di}{dt} = \lim_{h \to 0} \frac{\frac{6t - 6(t+h)}{t(t+h)}}{h}

Multiply by the reciprocal 1h\frac{1}{h} and expand the numerator terms:

didt=lim⁡h→06t−6t−6hh⋅t(t+h)=lim⁡h→0−6hh⋅t(t+h)\frac{di}{dt} = \lim_{h \to 0} \frac{6t - 6t - 6h}{h \cdot t(t+h)} = \lim_{h \to 0} \frac{-6h}{h \cdot t(t+h)}

Step 3: Divide out hh and Evaluate the Limit as h→0h \to 0

didt=lim⁡h→0−6t(t+h)=−6t(t+0)=−6t2 A/s\frac{di}{dt} = \lim_{h \to 0} \frac{-6}{t(t+h)} = \frac{-6}{t(t+0)} = -\frac{6}{t^2}\text{ A/s}

Step 4: Calculate Inductor Voltage VL(t)V_L(t)

Multiply the derivative of current by inductance L=0.5L = 0.5:

VL(t)=L⋅didt=0.5⋅(−6t2)=−3t2 VoltsV_L(t) = L \cdot \frac{di}{dt} = 0.5 \cdot \left(-\frac{6}{t^2}\right) = -\frac{3}{t^2}\text{ Volts}

Step 5: Evaluate at t=2 st = 2\text{ s}

VL(2)=−3(2)2=−34=−0.75 VV_L(2) = -\frac{3}{(2)^2} = -\frac{3}{4} = -0.75\text{ V}

Engineering Conclusion

At t=2 secondst = 2\text{ seconds}, the induced back-EMF voltage is −0.75 V-0.75\text{ V}. In electrical engineering, Lenz's law dictates that the negative sign represents an opposing electromotive force: because the current is decreasing (didt<0\frac{di}{dt} < 0), the collapsing magnetic field in the inductor attempts to sustain current flow by generating a potential difference opposing the rapid decay.