In the last how they have written $X_L - X_C =R$

2 votes

Hmm, a less confusing way of asking the question would be "which is the correct phase difference?"

I was criticising the question, not your title. I assumed that there is only one possible answer, in which case identifying the many incorrect answers less efficient than identifying the single correct answer.

However, thinking about it again, there is a *range* of possible answers, so the question "which options are incorrect" is not so silly. I have revised my answer.

However, thinking about it again, there is a *range* of possible answers, so the question "which options are incorrect" is not so silly. I have revised my answer.

2 votes

Best answer

**Revised Answer**

$\frac12 \lt \tan\phi = \frac{X_L-X_C}{R} \lt 1$ (see below). Therefore $0.1476\pi \lt \phi \lt \frac14\pi=0.25\pi$. The only possible option is (C) $\phi=\frac16\pi=0.1666...\pi$. (A, B, D) are incorrect.

$X_L-X_C=\omega L - \frac{1}{\omega C} = \frac{\omega^2 LC - 1}{\omega C}$

$\omega^2=(\omega_0+\frac{R}{2L})^2=\omega_0^2+\frac{R}{L}\omega_0+\frac{R^2}{4L^2}$

$\omega^2 LC - 1=\omega_0^2 LC+RC\omega_0+\frac{CR^2}{4L} - 1=(\omega_0+\frac{R}{4L})CR$

because $\omega_0^2 LC=1$.

Therefore

$\tan\phi=\frac{X_L-X_C}{R}=\frac{\omega^2 LC - 1}{\omega CR}=\frac{\omega_0+R/4L}{\omega_0+R/2L}=\frac{\alpha+1}{\alpha+2}$

where $\alpha=\frac{4L\omega_0}{R}=\frac{4}{R}\sqrt{\frac{L}{C}}$.

$C, L, R \gt 0$ so $\alpha \gt 0$ hence $\frac12 \lt \tan\phi \lt 1$.

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