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Question: An inductive coil has an internal resistance of 20 $\Omega$. When an AC Voltage source with a frequency of 100 Hz is connected to the coil, the current lags the voltage by 30 degrees. What is the value of the inductance L.

My Work: I am unsure where to begin here. If we know that $\omega_d=\frac{1}{\sqrt{LC}}$ and we also know that $f=2\pi \omega_d$, then we can just find the capacitance and solve for L in the first equation and we will have our answer. However, I am not sure how to find capacitance. Any help is appreciated.

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  • $\begingroup$ There seems to be no capacitance in this problem. Since they've mentioned an inductor, you can neglect it's capacitance. @Xittenn seems to have the right answer. $\endgroup$
    – Kitchi
    Commented Nov 12, 2012 at 18:45

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Given $\theta = 30$ and pure resistance is $20 \Omega$. Using a phase diagram it could be observed that $ \tan{\theta} = \frac{\chi_L}{R} $. Find $\chi_L$ and substitute that into $\chi_L = 2 \pi f L $. It's pretty straightforward I believe?

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  • $\begingroup$ Ok yes that makes sense at $\theta=30$, $X_C=0$. Thank Yuou $\endgroup$
    – Mathstudent
    Commented Nov 12, 2012 at 18:41
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Your capacitance formula is not applicable here. If you can work with complex numbers, you can use the impedance of the inductor, $j \omega L$, to calculate the total impedance ($Z = v/i = R_{internal}+ j \omega L$), and then work out what L must be to yield the specified phase angle at the specified frequency.

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Unless I'm mistaken you might wish to check out Inductive Reactance It's straight substitution

An inductor causes a lag of $\pi/2$; so when the lag is 30 degrees and given the internal resistance - what remains is the inductance.

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  • $\begingroup$ I thought about that but i'm not sure how to find $X_L$ $\endgroup$
    – Mathstudent
    Commented Nov 12, 2012 at 18:27

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