Question

When the current coil changes from $5\mathrm{A}$ to $2\mathrm{A}$ in $0.1\mathrm{s}$, an average of $50\mathrm{V}$ is produced. The self-inductance of the coil is.

• A. $6\mathrm{H}$
• B. $0.67\mathrm{H}$
• C. $1.67\mathrm{H}$
• D. $3\mathrm{H}$

Answer 1

The correct option is C

$1.67\mathrm{H}$

Step 1: Given data

$\mathrm{E}=50\mathrm{V}$

Time, ${\mathrm{d}}_{\mathrm{t}}=0.1\mathrm{s}$

The inductor resists changes in current; as the current decreases, the inductor converts magnetic energy to electrical energy by creating a potential difference.

Step 2: Formula used

The voltage drop across the inductor is calculated as follows:

$\mathrm{V}=\mathrm{L}\frac{{\mathrm{d}}_{\mathrm{i}}}{{\mathrm{d}}_{\mathrm{t}}}$

Step 3: Calculation

By putting the values in the formula, we get

$$\begin{gathered} 50=-\frac{\mathrm{L}(2–5)}{0.1} \\ 50=-\mathrm{L}(-30) \\ 50=30\mathrm{L} \\ \mathrm{L}=\frac{50}{30} \\ \mathrm{L}=1.67\mathrm{H} \end{gathered}$$

Therefore the correct option is C.