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A 1.07 H inductor is connected in series with a fluorescent lamp to limit the current drawn by the lamp. If the combination is connected to a 28.9 Hz, 170 V line, and if the voltage across the lamp is to be 17.6 V, what is the current in the circuit

Respuesta :

asuwafohjames

Answer:

0.784 A

Explanation:

From the question,

Note that the current in the circuit is the same as the current flowing through the inductor since they are both connected in series.

I = VL/XL....................... Equation 1

Where I = current flowing through the circuit, VL = Voltage drop across the inductor,  XL = reactance of the inductor.

XL = 2πfL................. Equation 2

Given: f = 28.9 Hz, L = 1.07 H, π = 3.143

XL = 2(3.143)(28.9)(1.07)

XL = 194.38 Ω.

VL = V-Vf

VL = 170-17.6

VL = 152.4 V

Substitute these values into equation 1

I = 152.4/194.38

I = 0.784 A

The current in the circuit when combination is connected  should be 0.784 A.

Calculation of the current:

SInce

we know that

I = VL/XL....................... Equation 1

Here,

I = current flowing through the circuit,

VL = Voltage drop across the inductor,  

XL = reactance of the inductor.

And,

XL = 2πfL................. Equation 2

Here

f = 28.9 Hz, L = 1.07 H, π = 3.143

So,

XL = 2(3.143)(28.9)(1.07)

XL = 194.38 Ω.

Now

VL = V-Vf

VL = 170-17.6

VL = 152.4 V

Now

I = 152.4/194.38

I = 0.784 A

hence, The current in the circuit when combination is connected  should be 0.784 A.

Learn more about current here: https://brainly.com/question/11311946

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