Question

solve it for brainlist ..... ​

Answer 1

GIVEN :-

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

• It is connected to

• A 220V (rms) .
• 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" .

Answer 2

Answer:

The answer is 0.065 H

hope this helps you