# Which is the correct phase difference ?

107 views

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

edited May 10, 2017
Hmm, a less confusing way of asking the question would be "which is the correct phase difference?"
Thanks for the idea
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.

$\frac12 \lt \tan\phi = \frac{X_L-X_C}{R} \lt 1$ (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 $\frac12 \lt \tan\phi \lt 1$.

answered May 9, 2017 by (28,806 points)
selected May 10, 2017 by koolman
Then how in the last step they got $\phi =\pi /4$ which is correct accordinf to the answer
Sammy gerbil answer is excellent.since $X_L-X_C \neq R$ so $\phi \neq \pi/4$
The correct options should be $C$ and $D$ if it becomes a purely resistive circuit.
I think the circuit cannot become purely resistive. That is not consistent with there being a finite resonant frequency $\omega_0=\frac{1}{LC}$. So option (D) $\phi=0$ is not possible.