Question

If the current in a coil changes from \(5\,\text{A}\) to \(2\,\text{A}\) in \(0.1\,\text{s}\) and an average emf of \(30\,\text{V}\) is induced, find the self-inductance.

• A. \(0.5\,\text{H}\)
• B. \(1\,\text{H}\)
• C. \(2\,\text{H}\)
• D. \(3\,\text{H}\)

Hint:

For self-inductance problems remember the relation: \[ E = L\frac{\Delta I}{\Delta t} \] A larger rate of change of current produces a larger induced emf.

Answer 1

The Correct Option is B

Concept: The induced emf in a coil due to self-inductance is given by \[ E = L\frac{dI}{dt} \] where
• \(E\) = induced emf
• \(L\) = self-inductance
• \(\frac{dI}{dt}\) = rate of change of current

Step 1: Identify the given quantities. \[ E = 30\,\text{V} \] \[ I_1 = 5\,\text{A}, \quad I_2 = 2\,\text{A} \] \[ \Delta t = 0.1\,\text{s} \]

Step 2: Find the rate of change of current. \[ \frac{dI}{dt} = \frac{I_1 - I_2}{\Delta t} \] \[ = \frac{5 - 2}{0.1} \] \[ = \frac{3}{0.1} = 30\,\text{A/s} \]

Step 3: Substitute into the formula. \[ E = L\frac{dI}{dt} \] \[ 30 = L(30) \] \[ L = 1\,\text{H} \] Thus, \[ \boxed{L = 1\,\text{H}} \]