Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

Series inductor needed for an arc lamp to operate

2 votes
52 views

An arc lamp requires a direct current of 10A and 80V to function.If it is connected to 220 V (rms) ,50 Hz AC supply the series inductor needed for it to work is close to:

A) 0.08 H
B) 0.044 H
C) 0.065 H
D) 80 H

Since the frequency is less I thought that it is simply an output of a rectifier consisting of AC ripples.The DC component of the rectifier I thought was the average value of AC ripples which is $200 \sqrt 2 * \frac{2}{\pi}$ and the resistance of inductor is $2\pi f L$ So since inductor is connected in series to the load resistance current through inductor is 10A . Hence $$10=\frac{\text{average voltage or dc voltage}}{\text{inductor reactance}}$$ This gives me an answer of 0.065 H

However I am not sure the method I adopted is correct because I am taking capacitance reactance in DC voltage which should be zero.

asked Mar 20, 2018 in Physics Problems by Navin (280 points)
The question does not mention a DC rectifier, so you should not assume one is used. The question does not explain how the arc lamp works, but I would assume that a minimum amount of power is required. What inductance is required to deliver the same power?

1 Answer

1 vote
 
Best answer

The question does not mention a DC rectifier, so you should not assume one is used. The question does not explain how the arc lamp works, but I would assume that it is a resistor and that a minimum amount of power is required for it to work.

Using DC, the arc lamp dissipates a power of $P=I^2 R$. Its resistance is $R=\frac{80V}{10A}=8\Omega$.

Using AC the arc lamp dissipates a power of $P=I_{rms}^2 R$. The power dissipated will be the same as when using DC provided that $I_{rms}=I=10A$, because the resistance remains the same.

The impedance of the RL circuit is
$|Z|=\frac{V_{rms}}{I_{rms}}=\frac{220V}{10A}=22\Omega$

$Z$ is related to $R, \omega, L$ by
$Z=R+j\omega L$
$|Z|^2=R^2+(\omega L)^2$
Therefore
$\omega L=\sqrt{|Z|^2-R^2} \approx 20.494 \Omega$
$L\approx \frac{20.494}{2 \pi 50} \approx 0.06523 H$

Answer : C.

answered Apr 24, 2018 by sammy gerbil (27,556 points)
edited Apr 27, 2018 by sammy gerbil
...