Question

Show that Equation$32.27$for the phase angle$\varphi$of a series$RLC$ circuit gives the correct result for a capacitor-only circuit.

Short answer

The expression $\varphi ={\tan }^{-1}\left(\frac{X_L-X_C}{R}\right)$for phase angle of series $RLC$circuit gives correct result for capacitor only circuit.

Step-by-step solution

Given Information

We are given that the Equation $32.27$which is $\varphi ={\tan }^{-1}\left(\frac{X_L-X_C}{R}\right)$.

We need to prove that this equation for phase angle of series $RLC$circuit gives correct result for capacitor only circuit.

Explanation 

We know equation,

$\tan \varphi =\frac{V_L-V_C}{V_R}\Rightarrow \frac{\left(X_L-X_C\right)}{RI}$

Here gets canceled so equation can be rewritten as :

$\varphi ={\tan }^{-1}\left(\frac{X_L-X_C}{R}\right)$

We are also given that , $X_L=X_C=0$

From here phase angle becomes$\varphi ={\tan }^{-1}\left(0\right)=0rad.$

Similarly for $AC$Inductor circuit conditions are

localid="1649931138237" $$\begin{gathered} R=X_C=0 \\ \varphi ={\tan }^{-1}\left(\infty \right)=\mathrm{\pi }/2\mathrm{rad} \end{gathered}$$

Hence proved that it gives correct result for capacitor only circuit.