Resonance And Locus Diagrams Practice Questions

At resonance imaginary part of $Z_{eq}=0$ Real of $Z_{eq}=\frac{R_1 X_c^2}{R_1^{2+}X-c^2}$ $\;\;=\frac{100 \times 100 \times 100}{100^{2+}100^2}=50\Omega$

The resonant frequency for the circuit is $\omega _0=\frac{1}{\sqrt{LC}}\sqrt{\frac{R_L^{2-}L/C}{R_C^{2-}L/C}}$ Since, $(R_L=R_C=R)$ So the circuit will have zero real part of admittance when, $R=\sqrt{\frac{L}{C}}=\sqrt{\frac{0.02}{100\mu F}}=14.14\Omega$

Usingstar to delta conversion, Line current will be zero when the parallel pair of induction capacitor is resonant at f=50Hz So, $50 \times 2\pi=\frac{1}{\sqrt{LC/3}}$ $100 \pi=\frac{1}{\sqrt{LC/3}}$ Since, L=10mH C will be 3.03 mF.

$Z=j\omega L||\frac{1}{j\omega C}$ $\;\;=\frac{L/C}{j\omega L+\frac{1}{j\omega C}}$ $\;\;=\frac{(L/C)j\omega C}{-\omega ^2 LC+1}$ $Z=\frac{j\omega L}{1-\omega^2 LC}$ $|Z|=\frac{\omega L}{1-\omega^2 LC}$ For, $1\gt \omega^2 LC;$ $Z=+ve$ For, $1 \lt \omega^2 LC;$ $Z=-ve$

Given $\omega _1=1 k$ rad/s, $V_1=100\angle 0^{\circ}V,$ $I_1=0.03\angle 31^{\circ}A$ At, $\omega _2=2 k$ rad/s, $V_2=100\angle 0^{\circ}V,$ $I_2=2\angle 0^{\circ}A$ At $\omega _2=2 k$ rad/s, voltage and current are in phase. Thus, it is a case of series resonance, $X_{L_{\omega 2}}=X_{C_{\omega 2}}$ $\therefore \;\;Z=R=\frac{V_2}{I_2}$ $\;\;=\frac{100\angle 0^{\circ}}{2\angle 0^{\circ}}=50\Omega$ $\therefore$ Resistance of circuit, $R=50\Omega$ Now at $\omega _1=1 k$ rad/s, $Z=\frac{V_1}{I_1}=\frac{100\angle 0^{\circ}}{0.03\angle 31^{\circ}}$ $\;\; =\frac{100}{0.03}\angle 31^{\circ}\Omega\;\;...(i)$ Also, $Z=|Z|\angle \tan^{-1}\left[ \frac{X_L-X_C}{R} \right] \;\;...(ii)$ Comparing equation (i) and (ii), we have: $-31^{\circ}=\tan ^{-1}\left[ \frac{X_L-X_C}{R} \right]$ $\tan(-31^{\circ})=\frac{X_L-X_C}{R}$ $X_L-X_C=R \tan(-31^{\circ})$ $\;\;=50X-0.6=-30$ $\therefore \;\; X_{L_{\omega 1}}-X_{C_{\omega 2}}=-30\;\;...(iii)$ $X_{L_{\omega 2}}=X_{C_{\omega 2}}$ or $\omega _2 L=\frac{1}{\omega _2C}$ $L=\frac{1}{\omega _2^2C}\;\;...(iv)$ From equation (iii), $\omega _1L-\frac{1}{\omega _1C}=-30$ $\omega _1\left ( \frac{1}{\omega _2^2C}\right )-\frac{1}{\omega _1C}=-30$ $\frac{\omega _1}{\omega _2^2C}-\frac{1}{\omega _1C}=-30$ $\frac{1 \times 10^3}{4 \times 10^6 C}-\frac{1}{10^3C}=-30$ $\frac{10^{-3}}{4C}-\frac{10^{-3}}{C}=-30$ $\frac{-3}{4C} \times 10^{-3}=-30$ $C=\frac{3 \times 10^{-3}}{4 \times 30}$ Substituting the value of C in equation (iv), we get, $L=\frac{1}{\omega _2^2C}$ $\;\;=\frac{1}{(2 \times 10^3)^2 \times 25 \times 10^{-6}}$ $\;\; =\frac{1}{100}=10mH$ Therefore, values are: $R=50\Omega ,$ $L=10mH,$ $C=25\mu F$

$q_1=\frac{\omega L_1}{R_1}$ $L_1=\frac{q_1R_1}{\omega }$ Similarly, $L_2=\frac{q_2R_2}{\omega }$ $Q=\frac{\omega (L_1+L_2)}{R_1+R_2}$ $=\frac{q_1R_1+q_2R_2}{R_1+R_2}$

Input impedance $z=j\omega L+R||\frac{1}{j\omega C}$ $z=j\omega L+\frac{R}{1+j\omega RC}$ $\therefore \;\; z=j0.1\omega +\frac{1}{1+j\omega } \times \frac{1-j\omega }{1-j\omega }$ $\;\;=j0.1\omega +\frac{1-j\omega }{1+\omega ^2}$ $\;\;=\frac{1}{1+\omega ^2}+j \times \left ( 0.1\omega -\frac{\omega }{1+\omega ^2} \right )$ At resonance, imaginary part must be zero. $0.1\omega - \frac{\omega }{1+\omega ^2}=0$ $0.1=\frac{1 }{1+\omega ^2}$ $\omega ^{2+}1=10$ $\omega ^2=9$ $\omega =3 rad/sec$

Let ,capacitive reactance $=X_C$ $I=\frac{V\angle 0^{\circ}+V\angle 0^{\circ}}{R-jX_C}$ $\;\;=\frac{2V}{R-jX_C}$ Using KVL, $V_{YX}+IR-V=0$ $V_{YX}=V-IR$ $V_{YX}=V-\left ( \frac{2V}{R-jX_C} \right )R =\frac{V(R-jX_C)-2VR}{R-jX_C} =-\frac{V(R+jX_C)}{R-jX_C}$ Method-1: $V_{YX}=-V\left[ \frac{R+jX_C}{R-jX_C} \right]$ When R=0 $V_{YX}=-V\left[ \frac{0+jX_C}{0-jX_C} \right]=V$ $V_{YX}=-V \times \left[ \frac{1+j\frac{X_C}{R}}{1-j\frac{X_C}{R}} \right]$ When $R\rightarrow \infty$ $V_{YX}=-V$ Method-2: $V_{YX}=-V\left[ \frac{R+jX_C}{R-jX_C} \right]$ $\;\;=V\angle 180^{\circ} \times \left ( \frac{\sqrt{R^2 X_C^2}\angle \tan^{-1}\left ( \frac{X_C}{R} \right )}{\sqrt{R^2 X_C^2}\angle \tan^{-1}\left ( \frac{-X_C}{R} \right )} \right )$ $\;\;=V\angle \left ( 180^{\circ} +2 \tan ^{-1}\left ( \frac{X_C}{R} \right ) \right )$ Magnitude of $V_{YX}=V$ So, option (C) and (D) can not be correct, as magnitude is 2 V in these two options. Angle of $V_{YX}=180^{\circ}+2 \tan ^{-1}\left ( \frac{X_C}{R} \right )$ When, R=0 $\angle V_{YX}=180^{\circ}+2 \tan ^{-1}(\infty )$ $\;\;=180^{\circ}+2 \times 90^{\circ}=360^{\circ}$ when $R=\infty$ $\angle V_{YX}=180^{\circ}+2 \tan ^{-1}(0 )=180^{\circ}$ on the basis of above analysis, the locus of $V_{YX}$ is drawn below:

Admittance of the series connected RLC $Y=\frac{1}{R+j\left ( \omega L-\frac{1}{\omega C} \right )}$ $Y=\frac{R-j\left ( \omega L-\frac{1}{\omega C} \right )}{R^{2+}\left ( \omega L-\frac{1}{\omega C} \right )^2}$ Separating, real and imaginary part of admittance. $Re[Y]=\frac{R}{R^{2+}\left ( \omega L-\frac{1}{\omega C} \right )^2}$ For any value of $\omega$ , the real part is always positive. When, $\omega L=\frac{1}{ \omega C}$ At, $\omega_0=\frac{1}{\sqrt{LC}}$ $Re[Y]=\frac{1}{R}$ $I_m(Y)=\frac{-\left ( \omega L-\frac{1}{\omega C} \right )}{R^{2+}\left ( \omega L-\frac{1}{\omega C} \right )^2} =\frac{\left ( \frac{1}{\omega C}-\omega L \right )}{R^{2+}\left ( \omega L-\frac{1}{\omega C} \right )^2}$ At, $\omega_0=\frac{1}{\sqrt{LC}}$ Imaginary part of zero $\Rightarrow \;\; Im(Y)=0$ For, $0 \lt \omega \lt \omega _0$ $\frac{1}{\omega C} \gt \omega L$ Therefore, $Im[Y] \gt 0$ For, $\omega _0 \lt \omega \lt \infty$ $\frac{1}{\omega C}\lt \omega L$ Therefore, $Im[Y] \lt 0$ On the basis of above analysis, the admittance locus is

At resonance, $X_L=X_C$ So, $Z=R_A+R_B$ Therefore, input impedance in purely resistive, is minimum, and the input voltage and output current are in phase. So, $V_1$ and $I$ are in phase $V_2=\frac{V_1}{R_A+R_B+j(X_L-X_C)}\times [R_B+j(X_L-X_C)]$ But, $X_L=X_C$ $V_2=\frac{V_1}{R_A+R_B} \times R_B$ Therefore, $V_2$ is in phase with $V_1$ and $V_2 \lt V_1$ voltage across the capacitor $V_C=I \times X_C=I \times \frac{1}{j\omega C}$ $V_C=\frac{1}{\omega C}\angle -90^{\circ}$ So, $V_C$ lags the current by $90^{\circ}.$ The phasor diagram on the basis of above analysis.

Atresonance, the circuit should be in unity power factor. Hence, 'Z' should be capacitive. Admittance of the parallel circuit. $Y=\frac{1}{jL\omega }+\frac{1}{1/jC\omega }=0$ $\frac{-1}{L\omega }+C\omega =0$ $\therefore \;\; C=\frac{1}{L\omega ^2}=\frac{1}{2(2 \pi \times 500)^2} =0.05\mu F$

$V=V_R+i(V_L-V_C)$ Since, $|V_L|=|V_C|$ and $|V_L|=2|V_R|$ Therefore, the circuit is at resonance and $V_R=V$ Quality factor $=\frac{V_L}{V}=\frac{V_L}{V_R}=\frac{2V_R}{V_R}=2$ As we know, $Q=\frac{\omega _0L}{R}$ $\Rightarrow \;\; 2=\frac{2 \pi f \times L}{5}$ $\Rightarrow \;\;L=31.8mH$

$\omega _0=\frac{1}{\sqrt{LC}}$ $\;\;=\frac{1}{\sqrt{100 \times 10^{-6} \times 10^{-6}}}=10^5 r/s$ $\Delta \omega =\frac{R}{L}=\frac{50}{100 \times 10^{-6}}$ $\;\;=50 \times 10^4 r/s$ $\omega _{lower}=\left[ \sqrt{\omega _0^{2+}\left ( \frac{\Delta \omega }{2} \right )^2}- \frac{\Delta \omega }{2}\right]$ $\;\;=\left[ \sqrt{(10^5)^{2+}\left ( \frac{5 \times 10^5}{2} \right )^2}-\frac{5 \times 10^5}{2} \right]$ $\;\;=10^5\left[ \sqrt{1+6.25}-2.5 \right]$ $\;\;=0.193 \times 10^5 r/s$ Hence, $f_{lower}=\frac{\omega _{lower}}{2 \pi}$ $\;\;=\frac{0.193 \times 10^5}{2 \pi}$ $\;\;=3065 Hz\approx 3.055kHz$

$Y=j\omega C+\frac{1}{30\angle 40^{\circ}}$ $\;\;=j\omega C+0.0255-j0.0214$ $\;\;=0.0255+j(\omega C-0.0214)$ $\;\;=Real(Y)+k Img(Y)$ To have unity power factor at ac source i.e. resonance condition, $Img(Y)=0$ $\Rightarrow \;\omega C-0.0214=0$ $\because \; \omega =2 \pi \times 50$ $\therefore \; C=\frac{0.0214}{100 \pi}=68.1\mu F$

In a series RLC circuit, at resonance $V_L=jQV_{source}$ and $V_c=-jQV_{source}$ also for $Q \gt 1, |V_c| \gt |V_{source}|$ Hence, option (C) is correct.