A current is made of two components, a DC component of `I_(1)=3amp` and an AC component given by `I_(2)=4sqrt2 sin omegat amp`, thenthe reading of th hot wire ammetero is : (Read the RMS value)

Explanation

`i=i_(1)+i_(2)=3+4sqrt2 sin omegat`
`i_("rms")^(2)=(int_(0)^(T)i^(2).dt)/(int_(0)^(T)dt)=(int_(0)^(T)(3+4sqrt2sin omega t)^(2)dt)/(T)`
`i_("rms")^(2)=(1)/(T)int_(0)^(T)(9+24sqrt2sin omega t+32 sin^(2)omegat)dt`
`therefore i_("rms")=5A`