Question

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.

Answer 1

As for lamp `V_(R)=IR=10xx5=50 V`,so when it is connected to `160 V` `ac` source through a choke in series.
`V^(2)=V_(R)^(2)+V_(L)^(2) , V_(L)=sqrt(160^(2)-50^(2))=152 V`
and as, `V_(L)=IX_(L)=IomegaL=2pifLI`
So, `L=V/(2pifI)=152/(2xxpixx50xx10)=4.84xx10^(-2)H`
Now the lamp is to be operated at `160 V dc`, instead of choke if additional resistance `r` is put in series with it
`V=I(R+r)`,i.e.`160=10(5+r)`
i.e.,`r=11 Omega`
In case of `ac` ,as choke has no resistance , power loss in the choke will be zero while the bulb will consume,
`P=I^(2)R=10^(2)xx5=500 W`
However, in case of `dc` as resistance `r` is to be used instead of choke the power loss in the resistance `r` will be `PL=10^(2)xx11=1100 W`
while the bulb will still consume `500 W`, i.e. when the lamp is run on resistance `r` instead of choke more than double the power consumed by the lamp is wasted by the resistance `r`.