Question

In a coil when current changes from $10\mathrm{A}$ to $2\mathrm{A}$ in time $0.1\mathrm{s}$, induced emf is $3.28\mathrm{V}$. What is self-inductance of the coil?

• A. $4\mathrm{H}$
• B. $0.4\mathrm{H}$
• C. $0.04\mathrm{H}$
• D. $5\mathrm{H}$

Answer 1

The correct option is C

$0.04\mathrm{H}$

Step 1: Given data

The current changes from $10\mathrm{A}$ to $2\mathrm{A}$ is $\mathrm{dI}=10\mathrm{A}-2\mathrm{A}=8\mathrm{A}$

The time is $\mathrm{t}=0.1\mathrm{s}$

The induced emf is $3.28\mathrm{V}$

Step 2: Formula used

The induced emf, $\mathrm{emf}=\mathrm{L}\frac{\mathrm{dI}}{\mathrm{dt}}$

Where, $\mathrm{L}$ is self inductance and $\mathrm{I}$ is current

Step 3: Self-inductance of the coil

The induced emf, $\mathrm{emf}=\mathrm{L}\frac{\mathrm{dI}}{\mathrm{dt}}$

$$\begin{gathered} \mathrm{emf}=\mathrm{L}\frac{\mathrm{dI}}{\mathrm{dt}} \\ 3.28=\mathrm{L}\frac{8}{0.1} \\ 3.28=\mathrm{L}\times 80 \\ \mathrm{L}=\frac{3.28}{80} \\ \mathrm{L}=0.041\mathrm{H} \end{gathered}$$

Therefore, the self-inductance of the coil is $0.04\mathrm{H}$.