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A Choke coil is needed to operate an arc...

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.

Text Solution

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We know that`i_(rms)=(V_(rms))/(sqrt(R^(2)+omega^(2)L^(2)))`[in series RL AC circuit)
`therefore10=(160)/(sqrt(25+(2pixx50xxL)^(2)))`
`or25+(100piL)^(2)=16^(2)`
`implies25+10^(5)L^(2)=256impliesL=sqrt(231)/(10^(5))=0.05H`Let `R_(A)` be the additional resistance required for operating the arc lamp with 160-V DC source, then `10=sqrt((160)/(R_(A)+5))impliesR_(A)+5=16`
`thereforeR_(A)=16-5=11Omega`
AC power consumed by the arc lamp =`V_(rms)I_(rms)cosphi=P_(ac)`
wherecosphi`=(R)/(Z)=(R)/(sqrt(R^(2)+(omegaL)^(2)))=(5)/(16)`
`impliesP_(ac)=160xx10xx(5)/(16)=500W`
Now, DC power consumed by arc lamp`P_(dc)=160xx10=1600W`
Power loss = 1600 - 500 = 1100 W = 68.75%
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