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Questions  >>  CBSE XII  >>  Physics  >>  Alternating current
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Q)

A coil has an inductance of $5\;H$ and resistance $20\; \Omega$ . An emf of $100\; v$ is applied to it. What is the energy stored in the magnetic field, when the current has reached its final value .

$\begin{array}{1 1} 100\;J \\ 125\;J \\ 200\;J \\ 225\;J \end{array} $

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A)
Solution :
given $L= 5\;H$
$R= 20 \Omega$
$E_v= 100 \;v$
When current reaches its final value $L=0$
$I_0 = \large\frac{E_0}{R}$
$\qquad = \large\frac{\sqrt 2 E_v}{R}$
$\qquad= \large\frac{100 \sqrt 2}{20}$
$\qquad= 5 \sqrt 2 \;A$
Energy stored in magnetic field $= \large\frac{1}{2}$$ L I_0^2$
$\qquad= \large\frac{1}{2} $$ \times 5(5 \sqrt 2)^2$
$\qquad= 125\;J$
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