An ac source of voltage $\displaystyle \nu=\nu_{\mathrm{m}} \sin \omega \mathrm{t}$ is connected to a series combination of LCR circuit. Draw the phasor diagram. Using it obtain an expression for the impedance of the circuit and the phase difference between applied voltage and the current.

$\displaystyle V_{R m}=i_{m} R, V_{C m}=i_{m} X_{C}, V_{L m}=i_{m} X_{L}$ From phasor diagram: $\displaystyle V_{m}^{2}=V_{R m}^{2}+\left(V_{C m}-V_{L m}\right)^{2}$ $\displaystyle V_{m}^{2}=\left(i_{m} R\right)^{2}+\left(i_{m} X_{C}-i_{m} X_{L}\right)^{2}$ $\displaystyle =i_{m}^{2}\left[R^{2}+\left(X_{C}-X_{L}\right)^{2}\right]$ Or $\displaystyle i_{m}=\frac{V_{m}}{\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}}$ $\displaystyle \therefore i_{m}=\frac{V_{m}}{Z}$ $\displaystyle \therefore Z=\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}$ From phasor diagram: $\displaystyle \tan \theta=\frac{V_{C m}-V_{L m}}{V_{R m}}$ $\displaystyle =\frac{X_{C}-X_{L}}{R}$ $\displaystyle \therefore \theta=\tan ^{-1}\left(\frac{X_{C}-X_{L}}{R}\right)$