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 10 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

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

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Answer 2

Thus Power loss in the resistance r will be PL= 10^2 × 11 = 1100 W and the bulb will consume 500 watt.

Explanation:

As for lamp VR = IR = 10 × 5 = 50 V

So when it is connected to 160 V ac source through a choke in series.

V2 = V^2R + V^2L

VL = √160^2 − 50^2 = 152 V

and as, VL  = IXL= IωL = 2πfLI

So, L = V / 2πfI = 152 / 2 × π × 50 × 10

L = 4.84 × 10^−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 = 11Ω

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 × 5 = 500 W

Thus Power loss in the resistance r will be PL= 10^2 × 11 = 1100 W and the bulb will consume 500 watt.