Question

An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220V(rms),
50 Hz AC supply, the series inductor needed for it to work is close to :
(1) 0.08 H
(2) 0.044 H
(3) 0.065 H
(4) 80 H​

Answer 1

Answer:

Given that arc lamp requires a direct current of 10 A at 80 V to work efficiently. Applying Ohm's law we get the resistance of the lamp to be:

⇒ V = IR

⇒ R = V / I

⇒ R = 80 V / 10 A

⇒ R = 8 ohms

Now it is said that an inductor is conducted in series with 220 V (rms) source having frequency of 50 Hz. Hence it now turns out to be a L-R circuit.

We know that, in a L-R circuit:

⇒ V = IZ  ( Z = Impedance )   ... Eqn. (1)

Calculating Impedance we get:

⇒ Z = √ [ R² + ( wL )² ]

⇒ Z = √ [ 8² + ( 2 × 3.14 × 50 × L )² ]

⇒ Z = √ [ 64 + 98596(L²) ]

Substituting this in Eqn. (1) we get:

⇒ 220 = 10 {√ [ 64 + 98596(L²) ]}

⇒ 220/10 = √ [ 64 + 98596(L²) ]

⇒ 22 = √ [ 64 + 98596(L²) ]

Squaring on both sides we get:

⇒ 484 = 64 + 98596 L²

⇒ 484 - 64 = 98596 L²

⇒ 420 = 98596 L²

⇒ L² = 420 / 98596

⇒ L² = 0.00425

⇒ L = √0.00425

⇒ L = 0.0651 ≈ 0.065 H

Hence option (C) is the right answer.

Answer 2

GIVEN :-

• An arc lamp requires a direct current of 10A at 80V to function .

• It is connected to

1. A 220V (rms) .
2. And 50 Hz AC supply .

TO FIND :-

• The series inductor needed for the arc lamp .

SOLUTION :-

☯︎ Resistance of lamp(By Ohm's Law) is,

$\huge\blue\star$ $\sf\pink{R\:=\:\dfrac{V}{I}\:}$

➪ R = 80/10

➪ R = 8Ω

✪︎︎︎ When inductor is connected in series and source is applied,

$\huge\green\star$ $\sf\pink{V\:=\:I\:Z\:}$

✞︎ Where,

• $\bf\red{Z\:=\:\sqrt{R^2\:+\:(\omega\:L)^2}\:}$

$\bf{:\implies\:V\:=\:I\:\sqrt{R^2\:+\:(\omega\:L)^2}\:}$

✞︎ Where,

• $\bf\red{V}$ = 220 V

• $\bf\red{I}$ = 10 A

• $\bf\red{R}$ = 8 Ω

• $\bf\red{\omega}$ = 2 π f

$\bf{:\implies\:V\:=\:I\:\sqrt{R^2\:+\:(2\:\pi\:f\:L)^2}\:}$

✞︎ Where,

• $\bf\red{f}$ = 50 Hz

$\rm{:\implies\:220\:=\:10\:\sqrt{8^2\:+\:(2\times{\pi}\times{50}\times{L})^2}\:}$

$\rm{:\implies\:22\:=\:\sqrt{64\:+\:(100{\pi}\times{L})^2}\:}$

$\rm{:\implies\:(22)^2\:=\:64\:+\:(100{\pi}\times{L})^2\:}$

$\rm{:\implies\:484\:=\:64\:+\:(100{\pi}\times{L})^2\:}$

$\rm{:\implies\:484\:-\:64\:=\:(100{\pi}\times{L})^2\:}$

$\rm{:\implies\:420\:=\:(100{\pi}\times{L})^2\:}$

$\rm{:\implies\:\sqrt{420}\:=\:100{\pi}\times{L}\:}$

$\rm{:\implies\:20.49\:=\:100{\pi}\times{L}\:}$

$\rm{:\implies\:L\:=\:\dfrac{20.49}{100\times{3.14}}\:}$

$\rm{:\implies\:L\:=\:\dfrac{20.49}{314}\:}$

$\bf\blue{:\implies\:L\:=\:0.0652\:H\:\approx\:0.065\:H\:}$

$\huge\therefore$ The series inductor needed for the lamp is "0.065 H" .