Question

Current in a circuit falls from $5.0\mathrm{A}$ to $0.0\mathrm{A}$ in $0.1\mathrm{s}$. If an average emf of $200\mathrm{V}$ is induced, calculate the self-inductance of the circuit.

Answer 1

Step 1: Given information

Initial current, ${\mathrm{I}}_1=5.0\mathrm{A}$

Final current, ${\mathrm{I}}_2=0\mathrm{A}$

Average emf, $\mathrm{e}=200\mathrm{V}$

Time is taken for the charge, $\mathrm{t}=0.1\mathrm{s}$

Step 2: Finding the self-inductance of the circuit

According to the given data,

Change in current,

$\mathrm{dI}={\mathrm{I}}_1–{\mathrm{I}}_2$

By substituting the values,

$$\begin{gathered} \mathrm{dI}=5.0–0 \\ =5\mathrm{A} \end{gathered}$$

The relationship between self-inductance of the coil (L) and average emf can be expressed as,

$\mathrm{e}=\mathrm{L}\frac{\mathrm{dI}}{\mathrm{dT}}$

now,

$$\begin{gathered} \mathrm{L}=\frac{\mathrm{e}}{{\displaystyle \frac{\mathrm{dI}}{\mathrm{dt}}}} \\ =\frac{200}{{\displaystyle \frac{5}{0.1}}} \\ =4\mathrm{H} \end{gathered}$$

Therefore, the self-inductance of the circuit will be $4\mathrm{H}$.