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. Numerical answer type. Work it out, then open the answer.

Answer: 4. dI/dt = d/dt(3t+8) = 3 A/s |e| = L dI/dt ⇒ L = 12×10⁻³3 = 4×10⁻³ H So the self-inductance is 4 mH.

The current in an inductor is given by I = (3t + 8) where t is in… | JEE Main 2024 (5 Apr, Shift 2)

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Self-Inductance & Energy

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 $12\text{ mV}$ . The self-inductance of the inductor _________ $\text{mH}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 4

$\dfrac{dI}{dt} = \dfrac{d}{dt}(3t+8) = 3\text{ A/s}$
$|e| = L\,\dfrac{dI}{dt} \Rightarrow L = \dfrac{12\times10^{-3}}{3} = 4\times10^{-3}\text{ H}$
So the self-inductance is 4 mH.

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