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
A
1 rad/s
B
2 rad/s
C
3 rad/s
D
4 rad/s
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.