A Choke coil is needed to operate an arc lamp at `160 V` ("rms") and `50 Hz`. The lamp has an effective resistnce of `5 Omega` when running at 1`0 A("rms")`. Calculate the inductance of the choke coil. If the same arc lamp is to be operated on `160 V (DC)`, what additional resistance is required ? Compare the power loses in both cases.

Explanation

For lamp,
`(V_("rms"))_R=(i_("rms"))(R)=10xx5=50V`

`V=V_0sin omega t`
In series,
`(V_("rms"))^2=(V_("rms"))_R^2+(V_("rms"))_L^2`
`:. (V_("rms"))_L=sqrt((V_("rms"))^2-(V_("rms"))_R^2)`
`=sqrt((160)^2-(50)^2)`
`=152 V`
`As (V_("rms"))L=(i_("rms"))X_L=((i_("rms"))(2pifL)`
`:. L=((V_("rms"))_L)/((2pif)(i_("rms"))`
Substituting the values we get
`L=152/((2pi)(50)(10))`
`=4.84xx10(5+R')`
Now when the lamp is operated at `160 V , DC` and instead of choke let an additional resistance `R'` is put in series with it, then
`V=i(R+R)`
or `160=10(5+R')`
`:. R'=11Omega`
In case of AC, as the choke has no resistance , power loss in choke is zero.
In case of `DC`, the loss in additional resistance `R'` is
`P=i^2R'=(10)^2(11)`
`=1100W`.