Previous Year Questions- Resonance and Locus Diagrams | Network Theory (Electric Circuits) - Electrical Engineering (EE) PDF Download

Q1: The voltage v(t) across the terminals a and b as shown in the figure, is a sinusoidal voltage having a frequency ω = 100 radian/s. When the inductor current i(t) is in phase with the voltage v(t), the magnitude of the impedance Z (in Ω) seen between the terminals a and b is ________ (up to 2 decimal places).        (2018)
(a) 25
(b) 50
(c) 100
(d) 150
Ans: (b)
Sol: At resonance imaginary part of Z_eq = 0

Q2: 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) ?      (2018)
(a) (b) (c) 

(d) Ans: (b)
Sol: 
Q3: 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 ___________.     (SET-2 (2016))
(a) 1.32
(b) 2.12
(c) 3.03
(d) 4.08
Ans: (c)
Sol: Usingstar to delta conversion,
Line current will be zero when the parallel pair of induction capacitor is resonant at f = 50Hz
Since, L = 10mH
C will be 3.03 mF.

Q4: The circuit below is excited by a sinusoidal source. The value of R, in Ω, for which the admittance of the circuit becomes a pure conductance at all frequencies is _____________.       (SET-1  (2016))
(a) 14.14
(b) 8.62
(c) 22.46
(d) 12.18
Ans: (a)
Sol: The resonant frequency for the circuit is
So the circuit will have zero real part of admittance when,

Q5: 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       (SET-1 (2015))
(a) (b) (c) (d) Ans: (b)
Sol: 
Q6: A series RLC circuit is observed at two frequencies. At ω₁ = 1 krad/s, we note that source voltage V₁ = 100∠0° V results in a current  I₁ = 0.03∠31°A. At ω₂ = 2 krad/s, the source voltage V₂ = 100∠0° V results in a current I₂ = 2∠0^∘. The closest values for R, L, C out of the following options are        (SET-3 (2014))
(a) R = 50 Ω ; L = 25mH; C = 10 μ F;
(b) R = 50 Ω ; L = 10 mH; C = 25 μ F;
(c) R = 50 Ω ; L = 50 mH; C = 5 μ F;
(d) R = 50 Ω ; L = 5mH; C = 50 μ F;
Ans: (b)
Sol: Given
At ω₂ = 2k rad/s, voltage and current are in phase. Thus, it is a case of series resonance,
∴ Resistance of circuit,
Comparing equation (i) and (ii), we have:
From equation (iii),
Substituting the value of C in equation (iv), we get,
Therefore, values are:
$𝑅=50\Omega ,$R = 50Ω,
L=10mH,
C = 25μF  

Q7: Two magnetically uncoupled inductive coils have Q factors q₁ and q₂ at the chosen operating frequency. Their respective resistances are R₁ and R₂. When connected in series, their effective Q factor at the same operating frequency is        (2013)
(a) $𝑞1𝑅1+𝑞2𝑅2$q₁R₁ + q₂R₂
(b) $\left(𝑞1/𝑅1\right)+\left(𝑞2/𝑅2\right)$(q₁/R₁) + (q₂/R₂)
(c) $(𝑞1𝑅1+𝑞2𝑅2)/(𝑅1+𝑅2)$(q₁R₁ + q₂R₂)/(R₁ + R₂)  
(d) $𝑞1𝑅2+𝑞2𝑅1$q₁R₂ + q₂R₁ 
Ans: (c)
Sol: 
Q8: The resonant frequency for the given circuit will be    (2008)
(a) 1 rad/s
(b) 2 rad/s
(c) 3 rad/s
(d) 4 rad/s
Ans: (c)
Sol: Input impedance
At resonance, imaginary part must be zero.

Q9: 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     (2007)
(a) (b) (c) (d) Ans: (a)
Sol: Let, capacitive reactance = X_C
Using KVL,
Method-1:
When R = 0
Method-2:
Magnitude of V_YX  = V
So, option (C) and (D) can not be correct, as magnitude is 2 V in these two options.
on the basis of above analysis, the locus of V_YX is drawn below:

Q10: 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 ω is       (2007)
(a) (b) (c) (d) Ans: (d)
Sol: Admittance of the series connected RLC
Separating, real and imaginary part of admittance.
For any value of ω , the real part is always positive. When,
Imaginary part of zero
On the basis of above analysis, the admittance locus is

Q11:The circuit shown in the figure is energized by a sinusoidal voltage source V₁ at a frequency which causes resonance with a current of I.
The phasor diagram which is applicable to this circuit is      (2006)
(a) (b) (c) (d) Ans: (a)
Sol: At resonance,
Therefore, input impedance in purely resistive, is minimum, and the input voltage and output current are in phase.
So, V₁ and I are in phase
But, X_L = X_C
Therefore, V₂ is in phase with V₁ and V₂ < V₁ voltage across the capacitor
So, V_C lags the current by 90^∘.
The phasor diagram on the basis of above analysis.

Q12: The value of Z in figure, which is most appropriate to cause parallel resonance at 500 Hz is      (2004)
(a) 125.00 mH
(b)304.20 μF
(c) 2.0 μF
(d) 0.05 μF
Ans: (d)
Sol: Atresonance, the circuit should be in unity power factor.
Hence, 'Z' should be capacitive.
Admittance of the parallel circuit.

Q13: In the circuit of figure, the magnitudes of V_L and V_C are twice that of V_R. Given that f = 50 Hz, the inductance of the coil is      (2003)
(a) 2.14 mH
(b) 5.30 H
(c) 31.8 mH
(d) 1.32 H
Ans: (c)
Sol: Therefore, the circuit is at resonance and
V_R = V
 Quality factor As we know,

Q14: A series R-L-C circuit has R = 50Ω, L = 100μ and C = 1μ. The lower half power frequency of the circuit is       (2002)
(a) 30.55 kHz
(b) 3.055 kHz
(c) 51.92 kHz
(d) 1.92 kHz
Ans: (b)
Sol: 
Q15: In the circuit shown in figure, what value of C will cause a unity power factor at the ac source?      (2002)
(a) 68.1 μF
(b) 165 μF
(c) $0.681𝜇𝐹$0.681 μF
(d) 6.81 μF
Ans: (a)
Sol: To have unity power factor at ac source i.e. resonance condition,

Q16: In a series RLC circuit at resonance, the magnitude of the voltage developed across the capacitor     (2001)
(a) is always zero
(b) can never be greater than the input voltage
(c) can be greater than the input voltage, however, it is 90^∘ out of phase with the input voltage
(d) can be greater than the input voltage, and is in phase with the input voltage.
Ans: (c)
Sol: In a series RLC circuit, at resonance
Hence, option (C) is correct.