• # question_answer An inductor L, a capacitor C and a resistor R are connected in series in an AC circuit. Deduce with the help of suitable phasor diagram, a mathematical expression for impedance of this circuit. What is meant by resonance of this circuit? Prove that this circuit exhibits resonance at a frequency given by $1/2\pi \sqrt{LC}.$ OR  (i) A series L-C-R circuit is connected to an AC source of voltage V and angular frequency co when only the capacitor is removed, the current lags behind the voltage by a phase angle $'\phi '$ and when only the inductor is removed, the current leads the voltage by the same phase angle. Find the current flowing and the average power dissipated in the L-C-R circuit. (ii) An alternating voltage given by V=140 sin 314t is connected across a pure resistor of resistance 50 Q. Find (a) the frequency of the source (b) the rms current through the resistor.

Let an alternating voltage $V={{V}_{0}}\sin \omega t$ is applied across the series combination of L, C and R as shown in figure below. L-C-R circuit Since, I and ${{V}_{R}}$ are in same phase, ${{V}_{L}}$ leads I by phase $\pi /2$ and ${{V}_{C}}$ lags I by phase $\pi /2.$and So, ${{V}_{L}}$ and ${{V}_{C}}$ are in opposite phase and their resultant is $({{V}_{L}}-{{V}_{C}}).$ The phasor diagram is shown below, The resultant of ${{V}_{L}}$ and ${{V}_{C}}$ i.e. ${{V}_{L}}-{{V}_{C}}$ and ${{V}_{R}}$ must be equal to the applied voltage, given by the diagonal the parallelogram. By Pythagoras theorem,             ${{V}^{2}}=V_{R}^{2}+{{({{V}_{L}}-{{V}_{C}})}^{2}}$ But       ${{V}_{R}}=IR,$ ${{V}_{L}}=I{{X}_{L}},$ ${{V}_{C}}=I\,\,{{X}_{C}}$ where, ${{X}_{L}}=\omega L$ = inductive reactance ${{X}_{C}}=\frac{1}{\omega C}$ = capacitive reactance $\therefore$      ${{V}^{2}}={{(IR)}^{2}}+{{(I\,\,{{X}_{L}}-I\,\,{{X}_{C}})}^{2}}$ or,        $\frac{{{V}^{2}}}{{{I}^{2}}}={{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}$ or,        $\frac{V}{I}=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$ where, V and I are instantaneous voltage and current and V/I gives the net opposition offered by circuit which is known as impedance (Z) of circuit. $\therefore$      $Z=\sqrt{{{R}^{2}}+{{({{V}_{L}}-{{X}_{C}})}^{2}}}$ Resonance The series L-C-R circuit is said to be in resonance when the current through the circuit becomes maximum and inductive reactance becomes equal to the capacitance reactance i.e. $\therefore$      ${{X}_{L}}={{X}_{C}}$ or $\omega L=\frac{1}{\omega C}$ or,        ${{\omega }^{2}}=\frac{1}{LC}$ $\Rightarrow$   $\omega =\frac{1}{\sqrt{LC}}\Rightarrow 2\pi f=\frac{1}{\sqrt{LC}}$ where, f is resonant frequency, $f=\frac{1}{2\pi \sqrt{LC}}$ Hence proved. Or (i) As we know that, $\tan \phi =\frac{{{X}_{L}}-{{X}_{C}}}{R}=\frac{\omega L-\frac{1}{\omega C}}{R}$ Now, when capacitor is removed, $\tan \phi =\frac{\omega L}{R}$ and when inductor is removed, $\tan \phi =\frac{\omega C}{R}$ Negative sign indicates that current leads the voltage. $\therefore$      $\omega L=1/\omega C\Rightarrow \omega =1\sqrt{LC}$ $\Rightarrow L\text{-}C\text{-}R$ circuit is in resonance. $\therefore$      ${{i}_{rms}}={{V}_{eff}}/R=V/R$             ${{P}_{av}}={{V}_{rms}}{{i}_{rms}}={{V}^{2}}_{rms}/R=\frac{{{V}^{2}}}{R}$ (ii)  Given, $V=140\sin 314t,$ $R=50\Omega$ Comparing it with $V={{V}_{0}}\sin \omega t$ (a)        Here,     $\omega =314rad/s$ i.e.        $2\pi v=314$    $[\because \omega =2\pi v]$ $\Rightarrow$   $v=\frac{314}{2\pi }=\frac{31400}{2\times .314}=50Hz$ Frequency of AC, v = 50Hz             (b) As, ${{I}_{rms}}=\frac{{{V}_{rms}}}{R}$ and ${{V}_{rms}}=\frac{{{V}_{0}}}{\sqrt{2}}$             Here, ${{V}_{0}}=140V$ $\Rightarrow$   ${{V}_{rms}}=\frac{140}{\sqrt{2}}=70\sqrt{2}V$ $\therefore$      ${{I}_{rms}}=\frac{70\sqrt{2}}{R}=\frac{70\sqrt{2}}{50}=1.9A$ or 2A