Question

A 220V, 50Hz ac source is connected to an inductance of 0.2H and a resistance of `20Omega` in series . What is the current in the circuit?

Answer 1

rms value of current, `T_(rms)=E_(rms)/sqrt(R^2+(omegaL)^2)`
Here, ` E_(rms)=220V`,
`omega=2pif=2times3.14times50=314 Hz`
`sqrt(R^2+(omegaL)^2)=sqrt((20)^2+(314times0.2)^2)`
`therefore I_(rms)=220/(sqrt((20)^2+(314times0.2)^2))=3.34A`.