Solveeit Logo

Question

Physics Question on Alternating current

A coil of resistance 10Ω10\,\Omega and inductance 5H5\, H is connected to a 100V100\, V battery. Then the energy stored in the coil is

A

250J250\,J

B

250erg250\,erg

C

125J125\,J

D

125erg125\,erg

Answer

250J250\,J

Explanation

Solution

Final current, I=ERI=\frac{E}{R}
I=10AI=10\,A
Energy stored in the magnetic field
U=12LiU=\frac{1}{2}\,Li
12×5×(10)2=250J\frac{1}{2}\times5\times\left(10\right)^{2}=250\,J