Question

A choke coil is needed to oprated an arc lamp at 160V(rms)and 50Hz. The arc lamp has an effective resistance of 5 Omega when running of 10A(rms). Calculate the inductance of the choke coil. If the same arc lamp is to be operated on 160V(dc), what additional resistnce is requried? Compare the power losses in both the cases.

Answer 1

Explanation:

We know that

now please refer attachment 1

let Ra be the additional resistance required for operating the arc lamp with 160V DC source then,

now please refer attachment 2

AC power consumed by the arc lamp =Vrms Irms Cosϕ=Pac

now please refer attachment 3

Now DC power consumed by arc lamp

now please refer attachment 4

hope it helps you mark as brainliest please

Answer 2

Explanation:

As for lamp  V=iR= 10×5 =  50 V, So when it is connected to 160 v AC source through a choke in series

v²=$V_{R}$²+$V_{L}$²     v=√(160²-50²) = 152 V

and as

$V_{L}$ = $I$$X_{L}$ = iωL= 2∏fLI

so  L= $V_{L}$ / 2∏fi = 152 / 2∏×50×10 = 4.84×$10^{-2}$

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)

r = 11 ohm

In case of ac, as choke has no resistance, power loss in

the choke

will be zero while the bulb will consume,

P =  R = 100 × 5 = 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

power loss = $10^{2}$ × 11 = 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.

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