A coil of inductance 0:2 Henry and olmic resistance 40 ohm is connected to 300 V, 50 Hz a.c. supply. Find the maximum current in coil.

Explanation

`L= 2.0 H`
` R = 4Omega`
` E_v= 300 V`
`v = 50 Hz`
`Z=sqrt(R^2+(2pivL)^(2))`
`=sqrt((40)^2+(2xx3.14xx50xx0.2)^(2))`
`Z = sqrt(1600+(62.8)^(2))`
`=sqrt(1600+3943.84)`
`=sqrt(5543.84)`
`Z = 74.46Omega`
`I_v = E_v/2`
`I_v=300/(74.46)`
`I_v = 4.029A`
`I_v =I_0/sqrt2`
Maximum current in the coil `= sqrt2I_v`
`I_0= 1.414 xx4.029`
`I_0= 5.70 A`