When the current in an inductor coil is changed from \( 10\ \mathrm{A} \) to \( 25\ \mathrm{A} \) in a time interval of \( 1\ \mathrm{s} \), the self induced emf in the coil is \( 25\ \mathrm{V} \). What is the change in the energy stored in the inductor coil?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 437.5\ \mathrm{J} \)
The self-induced emf gives the self-inductance, \( |\varepsilon| = L\,\Delta I/\Delta t \), and the energy stored in an inductor is \( U = \frac{1}{2}LI^{2} \).
\( \displaystyle L = \frac{|\varepsilon|\,\Delta t}{\Delta I} = \frac{25\ \mathrm{V} \times 1\ \mathrm{s}}{(25 - 10)\ \mathrm{A}} = \frac{5}{3}\ \mathrm{H} \)
\( \displaystyle \Delta U = \frac{1}{2}L(I_{2}^{2} - I_{1}^{2}) = \frac{1}{2}\times\frac{5}{3}\times(625 - 100) \)
\( \Delta U = \frac{5}{6}\times525 = 437.5\ \mathrm{J} \)