A electric bulb of 100 w – 300 v is connected with an ac supply of 500 v and (150/) hz. the required inductance to save the electric bulb is
Answers
The required inductance to save the electric bulb is 4 Henry.
Explanation:
The voltage at which the bulb operate is 300 V
A.C supply has voltage 500 V
Therefore the difference of voltage VL=500−300 = 200 V
Will appear across the inductance offered by the coil of the bulb.
Therefore VL=ωL×I
I in this case is I=PV
=100500 = 15 A
Thus VL= 2πfL × I
200 = 2π × 150πL × 15
or L=20060 = 3.33 ≃ 4 Henry
Hence the required inductance to save the electric bulb is 4 Henry.
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A resistance of 9 ohm is connected to the terminal of a cell .a voltmeter connected across the cell reads is 1.8 volt on. when the resistance of 10 ohm in a series with 9 ohm the voltmeter reading changes to 1.9 volt calculate the EMF of the cell and its internal resistance ?
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Answer:
he correct option is C
Explanation:
The voltage at which the bulb operates is 300v And the aac supply has voltage 500v
Therefore, the difference of the voltage V
l
=500−300=200v
Will appear across the inductance of the spred by the coil of the bulb.
Therefore V
l
=ωL×I⇒
v
p
=
500
100
=
5
1
A.
Thus, V
l
=2πfL×I⇒200=2π×
π
150
l
5
1
⇒l=
60
200
=3.33≈4H