Question

A choke coil is needed to operate an arc lamp at 160V (RMS) and 50Hz. The arc lamp has an effective resistance of 5Ω when running at 10A (rms). Calculate the inductance of the choke coil. If the same arc lamp is to be operated on 160V DC, what additional resistance would be required. Compare the power losses in both the cases.

Answer 1

Step1:Given data:

$\raisebox{1ex}{$I$}\!\left/ \!\raisebox{-1ex}{$rms$}\right.$= 10

Step2: Formula used:

$\raisebox{1ex}{$I$}\!\left/ \!\raisebox{-1ex}{$rms$}\right.=$$\raisebox{1ex}{${\displaystyle \raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$rms$}\right.}$}\!\left/ \!\raisebox{-1ex}{$\sqrt{R^2+{\omega }^2{L}^2}$}\right.$

AC power consumed by the arc lamp=$\raisebox{1ex}{$V$}\!\left/ \!\raisebox{-1ex}{$rms{\displaystyle \raisebox{1ex}{$I$}\!\left/ \!\raisebox{-1ex}{$rms$}\right.}$}\right.\cos \varnothing =P_{ac}$

Calculation:

we know,$I_{rms}=\raisebox{1ex}{$V_{rms}$}\!\left/ \!\raisebox{-1ex}{$\sqrt{R^2+{\omega }^2}{L}^2$}\right.$

since,$\raisebox{1ex}{$I$}\!\left/ \!\raisebox{-1ex}{$rms$}\right.=10$

10 = $\raisebox{1ex}{$160$}\!\left/ \!\raisebox{-1ex}{$\sqrt{25+{(2\mathrm{\pi }\times 50\times \mathrm{L})}^2}$}\right.$

$25+{\left(100\mathrm{\pi L}\right)}^2$=${16}^2$

$25+{10}^5{L}^2=256$

$L=\sqrt{\raisebox{1ex}{$231$}\!\left/ \!\raisebox{-1ex}{${10}^5$}\right.}$

$=0.05H$

Let $R_a$ be additional resistance:

$10=\sqrt{\raisebox{1ex}{$160$}\!\left/ \!\raisebox{-1ex}{$R_a$}\right.}+5=16$

SINCE,$R_a=16-5=11$

AC power consumed :

$\cos \varnothing =\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$Z$}\right.=\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$\sqrt{R^2+{\omega }^2}{L}^2$}\right.$$=\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$16$}\right.$

$P_{ac}=160\times 10\times \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$16$}\right.=500W$

Step4: DC power consumed by arc lamp:

$P_{dc}=160\times 10=1600W$

Power loss$\frac{1100}{1600}\times 100=68.75\%$