The frequency of the voltage across the coil is 60 Hz.
The tendency of a coil to resist internal current fluctuations is known as self-inductance. Every time a coil's current varies, an EMF is produced that is inversely proportional to the rate of change of the coil's current.
Given the self-inductance (L) is 16 mH and resistance is 6.0 Ω.
L = 16×10⁻³ H
The frequency can be calculated using the inductive resistance formula, [tex]f=\frac{X_L}{2\pi L}[/tex] where, XL is inductive resistance and L is inductance.
First, we have to find inductive resistance using the phase angle formula.
[tex]\begin{aligned}\tan\phi&= \frac{X_L-X_C}{R}\\X_L&=R\tan\phi\\&=6\times \tan 45^{\circ}\\&=6\times1\\&=\mathrm{6\;\Omega}\end{aligned}[/tex]
Then, frequency is
[tex]\begin{aligned}f&=\frac{6}{2\pi \times 16\times 10^{-3}}\\&=\mathrm{60\;Hz}\end{aligned}[/tex]
The answer is 60 Hz.
The complete question is -
A coil with a self-inductance of 16 mH and a resistance of 6.0 Ω is connected to an ac source whose frequency can be varied. At what frequency will the voltage across the coil lead the current through the coil by 45°?
To know more about self-inductance:
https://brainly.com/question/28167218
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