L, E and i: 50.0mH Inductor and 24.0V AC Voltage Source

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A 50.0 mH inductor is connected to a 24.0V AC voltage source, with a current of 0.60A flowing through the circuit. The discussion focuses on determining the frequency of the voltage source using the relationship between inductance (L), voltage (E), and current (i). Impedance (z) is calculated as z = V/I, and the relationship z = Lω is established, leading to the equation ω = 2πf. The participants confirm the correct approach to relate these variables to find the frequency. The conversation emphasizes the importance of understanding impedance in AC circuits.
-Aladdin-
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A 50.0mh inductor is connected to an ac voltage source of 24.0V.The current flowing in this circuit is 0.60A.The frequency of the voltage source is :
a)50Hz
b)60Hz
c)1.27*10^3 Hz

My Work so far ! :
f = w/2pi.

But here we have L , E and i.

Any help will be welcomed.
 
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-Aladdin- said:
My Work so far ! :
f = w/2pi.

But here we have L , E and i.

Any help will be welcomed.
You can do better than this. How will you bring in L, E and i? Think impedance.
 
kuruman said:
You can do better than this. How will you bring in L, E and i? Think impedance.

So,impedance is : z=Lw = L*2pi*f ...

A guess , z = V/I = E/i ! but V(max)/I(max)
 
-Aladdin- said:
So,impedance is : z=Lw = L*2pi*f ...

A guess , z = V/I = E/i ! but V(max)/I(max)

If z = V/I, can you relate ω to V and I?
 
kuruman said:
If z = V/I, can you relate ω to V and I?

So is it , V/I=Lw then V/LI=w=2pi*f .
 
Correct.
 
kuruman said:
Correct.

Thank you, kuruman.
 
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