# Resonance and Locus Diagrams

 Question 1
The voltage v(t) across the terminals a and b as shown in the figure, is a sinusoidal voltage having a frequency $\omega$=100 radian/s. When the inductor current i(t) is in phase with the voltage v(t), the magnitude of the impedance Z (in $\Omega$) seen between the terminals a and b is ________ (up to 2 decimal places).
 A 25 B 50 C 100 D 150
GATE EE 2018   Electric Circuits
Question 1 Explanation:
At resonance imaginary part of $Z_{eq}=0$
Real of $Z_{eq}=\frac{R_1 X_c^2}{R_1^2+X-c^2}$
$\;\;=\frac{100 \times 100 \times 100}{100^2+100^2}=50\Omega$
 Question 2
A DC voltage source is connected to a series L-C circuit by turning on the switch S at time t=0 as shown in the figure. Assume i(0)=0, v(0)=0. Which one of the following circular loci represents the plot of i(t) versus v(t) ?

 A A B B C C D D
GATE EE 2018   Electric Circuits
Question 2 Explanation:

$I(s)=\frac{\frac{5}{s}}{s+\frac{1}{s}}$
$\;\;=\frac{5}{s^2+1}$
$i(t)=5 \sin t$
$v(t)=\frac{1}{C}\int_{0}^{t}i \; dt$
$v(t)=\int_{0}^{t}5 \sin t \; dt$
$v(t)=5[-\cos t]_0^t$
$v(t)=5[-\cos t+1]$
$v(t)=5-5 \cos t$

$\begin{matrix} t & i(t) & v(t)\\ 0 & 0 & 0\\ \frac{T}{4} & 5 & 5\\ \frac{T}{2}& 0 & 10\\ \frac{3T}{4} &-5 & 5\\ T& 0 & 0 \end{matrix}$
 Question 3
In the balanced 3-phase, 50 Hz, circuit shown below, the value of inductance (L) is 10 mH. The value of the capacitance (C) for which all the line currents are zero, in millifarads, is ___________.
 A 1.32 B 2.12 C 3.03 D 4.08
GATE EE 2016-SET-2   Electric Circuits
Question 3 Explanation:
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.
 Question 4
The circuit below is excited by a sinusoidal source. The value of R, in $\Omega$, for which the admittance of the circuit becomes a pure conductance at all frequencies is _____________.
 A 14.14 B 8.62 C 22.46 D 12.18
GATE EE 2016-SET-1   Electric Circuits
Question 4 Explanation:
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, $R=\sqrt{\frac{L}{C}}=\sqrt{\frac{0.02}{100\mu F}}=14.14\Omega$
 Question 5
An inductor is connected in parallel with a capacitor as shown in the figure.

As the frequency of current i is increased, the impedance (Z) of the network varies as
 A A B B C C D D
GATE EE 2015-SET-1   Electric Circuits
Question 5 Explanation:
$Z=j\omega L||\frac{1}{j\omega C}$
$\;\;=\frac{L/C}{j\omega L+\frac{1}{j\omega C}}$
$\;\;=\frac{(L/C)j\omega C}{-\omega ^2 LC+1}$
$Z=\frac{j\omega L}{1-\omega^2 LC}$
$|Z|=\frac{\omega L}{1-\omega^2 LC}$
For, $1\gt \omega^2 LC;$
$Z=+ve$

For, $1 \lt \omega^2 LC;$
$Z=-ve$
 Question 6
A series RLC circuit is observed at two frequencies. At $\omega _1$=1 krad/s, we note that source voltage $V_{1}=100\angle 0^{\circ}$ V results in a current $I_{1}=0.03 \angle 31 ^{\circ}A$. At $\omega _2$=2 krad/s, the source voltage $V_{2}=100\angle 0^{\circ}$ V results in a current $I_{2}=2 \angle 0^{\circ}$. The closest values for R, L, C out of the following options are
 A R = 50 $\Omega$ ; L = 25mH; C = 10 $\mu$ F; B R = 50 $\Omega$ ; L = 10 mH; C = 25 $\mu$ F; C R = 50 $\Omega$ ; L = 50 mH; C = 5 $\mu$ F; D R = 50 $\Omega$ ; L = 5mH; C = 50 $\mu$ F;
GATE EE 2014-SET-3   Electric Circuits
Question 6 Explanation:
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:
$R=50\Omega ,$
$L=10mH,$
$C=25\mu F$
 Question 7
Two magnetically uncoupled inductive coils have Q factors $q_1 \; and \; q_2$ at the chosen operating frequency. Their respective resistances are $R_1 \; and \; R_2$. When connected in series, their effective Q factor at the same operating frequency is
 A $q_{1}R_1+q_{2}R_2$ B $(q_{1}/R_1)+(q_{2}/R_2)$ C $(q_{1}R_{1}+q_{2}R_{2})/(R_{1}+R_{2})$ D $q_{1}R_{2}+q_{2}R_{1}$
GATE EE 2013   Electric Circuits
Question 7 Explanation:

$q_1=\frac{\omega L_1}{R_1}$
$L_1=\frac{q_1R_1}{\omega }$
Similarly,
$L_2=\frac{q_2R_2}{\omega }$

$Q=\frac{\omega (L_1+L_2)}{R_1+R_2}$
$=\frac{q_1R_1+q_2R_2}{R_1+R_2}$
 Question 8
The resonant frequency for the given circuit will be
GATE EE 2008   Electric Circuits
Question 8 Explanation:
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.
$0.1\omega - \frac{\omega }{1+\omega ^2}=0$
$0.1=\frac{1 }{1+\omega ^2}$
$\omega ^2+1=10$
$\omega ^2=9$
$\omega =3 rad/sec$
 Question 9
In the figure given below all phasors are with reference to the potential at point "O". The locus of voltage phasor $V_{XY}$ as R is varied from zero to infinity is shown by

 A A B B C C D D
GATE EE 2007   Electric Circuits
Question 9 Explanation:

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:
 Question 10
The R-L-C series circuit shown in figure is supplied from a variable frequency voltage source. The admittance-locus of the R-L-C network at terminals AB for increasing frequency $\omega$ is

 A A B B C C D D
GATE EE 2007   Electric Circuits
Question 10 Explanation:
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

There are 10 questions to complete.