A resistive coil of self-inductance 2 H and resistance `5 Omega` is connected across a battery of 50 V. How much energy will be stored in the inductor?

Options

A. 1000 joules
B. 10 joules
C. 100 joules
D. 500 joules

Answer

C

Explanation

Steady state current through the coil can be written as follows:
`i= E//R= 50//5= 10A`
Energy stored in the inductor is given by following formula:
`U= (1)/(2)Li^(2)`
`U= (1)/(2) xx2 xx10^(2) = 100 "joules"`
Hence, option (c) is correct.