Question

The current in an inductor is given by I = (3t + 8) where t is in second. The magnitude of induced emf produced in the inductor is 12 mV. The self inductance of the inductor _______ mH.

Answer 1

Correct Answer: 4

The induced electromotive force (\( \varepsilon \)) in an inductor is determined by its self-inductance (\( L \)) and the rate of change of current (\( \frac{dI}{dt} \)) according to the equation:

\[ |\varepsilon| = L \frac{dI}{dt}. \]

Step 1: Calculate the rate of change of current
Given the current \( I = 3t + 8 \). Differentiating with respect to time \( t \) yields:

\[ \frac{dI}{dt} = 3 \, \text{A/s}. \]

Step 2: Input known values
The magnitude of the induced emf is \( |\varepsilon| = 12 \, \text{mV} \), which is \( 12 \times 10^{-3} \, \text{V} \). Substituting into the formula:

\[ 12 \times 10^{-3} = L \cdot 3. \]

Step 3: Determine self-inductance (\( L \))
Rearranging the equation to solve for \( L \):

\[ L = \frac{12 \times 10^{-3}}{3} = 4 \times 10^{-3} \, \text{H}. \]

Converting to millihenries:

\[ L = 4 \, \text{mH}. \]

The self-inductance of the inductor is \( L = 4 \, \text{mH} \).