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Q. A coil of resistance $10\,\Omega$ and inductance $5\, H$ is connected to a $100\, V$ battery. Then the energy stored in the coil is

VITEEEVITEEE 2011

Solution:

Final current, $I=\frac{E}{R}$
$I=10\,A$
Energy stored in the magnetic field
$U=\frac{1}{2}\,Li$
$\frac{1}{2}\times5\times\left(10\right)^{2}=250\,J$