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A coil resistance $20 \Omega$ and inductance $5 \mathrm{H}$ is connected with a $100 \mathrm{~V}$ battery. Energy stored in the coil will be
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$62.50 \mathrm{~J}$
$U=\frac{1}{2} L i^2=\frac{1}{2} L\left(\frac{E}{R}\right)^2$ $=\frac{1}{2} \times 5 \times\left(\frac{100}{20}\right)^2=62.50 \mathrm{~J}$
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