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31. An inductance coil has a reactance of 100 P.
When an AC signal of frequency 1000 Hz is
applied to the coil, the applied voltage leads
the current by 45°. The self-inductance of the loss of heat to the surrounding) is close to
coil is [JEE (Main)-2020]?
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31. An inductance coil has a reactance of 100 P. When an AC signal of ...
The problem involves an inductance coil with given parameters. To determine the self-inductance, we will use the relationship between reactance, frequency, and inductance.
Given Data:
- Reactance (X_L) = 100 Ω
- Frequency (f) = 1000 Hz
- Voltage leads current by 45°
Understanding Reactance:
The inductive reactance (X_L) of a coil is given by the formula:
X_L = 2πfL
Where:
- X_L = inductive reactance
- f = frequency in Hz
- L = self-inductance in Henrys
Calculation of Self-Inductance:
We can rearrange the formula to solve for L:
L = X_L / (2πf)
Now, substituting the known values:
L = 100 / (2π × 1000)
Calculating this gives:
L = 100 / (2000π)
L ≈ 15.92 × 10^-3
L ≈ 15.92 mH
Conclusion:
Thus, the self-inductance of the coil is approximately:
L ≈ 15.92 mH
This value indicates the ability of the coil to store energy in the magnetic field created by the current flowing through it. The phase angle of 45° confirms that the circuit is behaving typically for an inductive load, where the voltage leads the current.
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31. An inductance coil has a reactance of 100 P. When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leadsthe current by 45°. The self-inductance of the loss of heat to the surrounding) is close tocoil is [JEE (Main)-2020]?
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