A `750 Hz,20 V` source is connected to a resistance of `100 Omega`, a capacitance of `1.0 muF` and an inductance of `0.18 H` in series.Calculate the following quantities:
(a)Impedence of the circuit
(b)Draw an impedence diagram with suitable scale
( c)Power factor
(d)The time in which the resistance will get heated by `10^(@)C`, provided that the thermal capacity of resistance =`2 J//^(@)C`

Explanation

`z=sqrt((X_(L)-X_(C))^(2)+R^(2))`
`omega=2npi=2(750)pi=1500 pi`
`X_(L)=omega_(L)=(1500pi)(0.18)=848.5 Omega`
`X_(C)=1/(omegaC)=10^(6)/(1500pi)=212.12 Omega`
`z=sqrt((636.4)^(2)+(100)^(2))`
`=100sqrt((6.364)^(2)+12)`
=`100x6.44~=644 Omega`
`tan phi=(X_(L)-X_(C))/R=(848.5-212.12)/100`
=`6.36`
( C)`cos phi=R/z=100/644=0.155`
(B)impedence is constant as `n` is constant
( D)`i_(rms)=epsilon_(rms)/z=20/644A`
`H=(i_(rms))^(2)RT=(20/644)^(2)(100)t`
`(ms)(Deltatheta)=(20/644)^(2) (100)t`
`(2)(10)=(20/644)^(2) (100)t`
`t=(644/20)^(2)xx1/100xx20`
`t=207.36 sec`