As for lamp `V_(R)=IR=10xx5=50V`, so when it si connected to 160V ac source through a choke in series.
`V^(2)=V_(R)^(2)+V_(L)^(2)`
`V_(L)=sqrt(160^(2)-50^(2))=152V`
`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 160V 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=11Omega `
In case of ac, as choke has no resitance, power loss in the choke will be zero while the bulb will consume.
`P=I^(2)R=10^(2)xx5=500W`
However, in case in dc as resistance r is to be used instead of choke, the power loss in the resistance r will be.
`PL=10^(2)xx11=1100W`
while the bulb will still consume 500W, i.e. whe 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.