Network Theory
Exam Duration: 45 Mins Total Questions : 30
In the circuit of the figure a charge of 600 C is delivered to the 100 V source in 1 min. the value of \({ V }_{ 1 }\) must be
- (a)
240 V
- (b)
120 V
- (c)
60 V
- (d)
30 V
In a series RLC circuit, R=2k\(\Omega\), L=1H and C=1/400 \(\mu\)F .The resonant frequency is
- (a)
\(2\times10^4 HZ\)
- (b)
\(\frac { 1 }{ \pi } \times 10^{ 4 }HZ\)
- (c)
\(10^{ 4 }HZ\)
- (d)
\(2\pi \times 10^{ 4 }HZ\)
A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q =100. If each of R, Land C is doubled from its original value, the new Q of the circuit is
- (a)
25
- (b)
50
- (c)
100
- (d)
200
A Lossy capacitor in series circuit model consists of R = 20\(\Omega\) and C = 25 pF. The equivalent parallel model at 50 kHz will have values of Rand C as
- (a)
1 G\(\Omega\), 10pF
- (b)
1.01 G\(\Omega\), 20pF
- (c)
810.5 \(\Omega\), 25pF
- (d)
810 G\(\Omega\), 25pF
In the circuit of figure I1 =4sin2tA and I2 =0
Find the value of V1
- (a)
-16 cos 2t V
- (b)
16 cos 2t V
- (c)
4 cos 2t V
- (d)
-4 cos 2t V
In the following non-planar graph number of independent loop equations are
- (a)
8
- (b)
12
- (c)
7
- (d)
5
The resistance of copper motor winding at t=20°C is 3.42\(\Omega\) After extended operation at full load, the motor winding measures 4.22\(\Omega\) If the temperature coefficient is 0.0426 then, what is the rise in temperature?
- (a)
600C
- (b)
45.20C
- (c)
72.90C
- (d)
10.160C
In a series resonant circuit, Vc =150 V, VL =150 V and VR =50 V. What is the value of the source voltage?
- (a)
Zero
- (b)
50 V
- (c)
350 V
- (d)
200 V
In a series RLC circuit at resonance, the magnitude of the voltage developed across the capacitor
- (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
A 50 μFcapacitor, when connected in series with a coil having 400 resistance, resonates at 1000 Hz.Find the inductance of the coil. Also, obtain the circuit current, if the applied voltage is 100 V. Also, calculate the voltage across the capacitor and the coiI at resonance.
- (a)
105.31V
- (b)
101.31V
- (c)
100.31V
- (d)
99.31V
A periodic rectangular signal X(t) has the waveform shown in the figure. Frequency of the fifth harmonic of its spectrum is
- (a)
40Hz
- (b)
200Hz
- (c)
250Hz
- (d)
1250Hz
For the circuit shown in figure.The Norton equivalent source current value is ....... and its resistance is
- (a)
1A, 3.5Ω
- (b)
2A, 4.5Ω
- (c)
1.5A, 4Ω
- (d)
2.5A, 3Ω
For the circuit shown in figure, the impedance function will have n numbers of pole of multiplicity 1 at ∞, where n is
- (a)
zero
- (b)
2
- (c)
3
- (d)
1
The curre~t having the waveform shown in figure is flowing in a resistance of 100.The average power is
- (a)
1000W
- (b)
1000/2W
- (c)
1000/3W
- (d)
1000/4W
An independent voltage source in series with an impedance Zs =Rs = jXs delivers a maximum average power to a load impedance ZL when
- (a)
ZL=Rs + jXs
- (b)
ZL=Rs
- (c)
ZL=jXs
- (d)
ZL=Rs-jXs
For the following circuit, the values of VTh and RTh are
- (a)
-100 V, 75 Ω
- (b)
155 V, 55 Ω
- (c)
155 V, 37Ω
- (d)
145 V, 75 Ω
In the following circuit a network and its Thevenin and Norton equivalents are given
The values of the parameters VTh, RTh, IN and RN are
- (a)
4V 2\(\Omega\) 2A 2\(\Omega\)
- (b)
4V 2\(\Omega\) 2A 3\(\Omega\)
- (c)
8V 1.2\(\Omega\) 30/3A 1.2\(\Omega\)
- (d)
8V 5\(\Omega\) 8/5A 5\(\Omega\)
Two elements are connected in series as shown in figure . Element 1 supplies 42 W of power Elements 2
- (a)
absorbs 72W
- (b)
absorbs 36W
- (c)
absorbs 63W
- (d)
absorbs 144W
For the three-phase circuit shown in figure, the ratio of the current IR:IY:IB is given by
- (a)
\(1:1:\sqrt { 3 } \)
- (b)
1:1:2
- (c)
1:1:0
- (d)
\(1:1:\sqrt { \frac { 3 }{ 2 } } \)
For the circuit shown in the figure, the value of R for critical damping will be
- (a)
10.5\(\Omega \)
- (b)
6\(\Omega \)
- (c)
3.5\(\Omega \)
- (d)
3\(\Omega \)
The input impedance Zin of the circuit shown in figure below is
- (a)
0.52 -j4.30\(\Omega \)
- (b)
0.52 + j15.70\(\Omega \)
- (c)
64.73 + j17.77\(\Omega \)
- (d)
0.3 - j33.66\(\Omega \)
A ramp voltage V(t) =100 V, is applied to an RC differentiating circuit with R = 5k\(\Omega \) and C = 4 \(\mu F\).The maximum output voltage is
- (a)
0.2 V
- (b)
2.0 V
- (c)
10.0 V
- (d)
50.0 V
For the given circuit, values of V1 and V2 respectively are
- (a)
0 and 5 V
- (b)
5 and 0 V
- (c)
5 and 5 V
- (d)
2.5 and 2.5 V
The two generator equivalent of a general network in terms of open-circuit impedance function is shown below.
The one generator equivalent is
- (a)
- (b)
- (c)
- (d)
If a two-port network is passive, then we have with the usual notation, the following relationship:
- (a)
\(Y=\left[ \begin{matrix} 0.2 & -0.1 \\ -0.1 & 0.2 \end{matrix} \right] \)
- (b)
\(Y=\left[ \begin{matrix} -0.2 & -0.1 \\ -0.1 & 0.2 \end{matrix} \right] \)
- (c)
\(Y=\left[ \begin{matrix} 0.2 & -0.1 \\ 0.1 & -0.2 \end{matrix} \right] \)
- (d)
\(Y=\left[ \begin{matrix} -0.2 & 0.1 \\ 0.1 & -0.2 \end{matrix} \right] \)
For the circuit shown in figure below, the value of RTh is
- (a)
100\(\Omega \)
- (b)
136.4\(\Omega \)
- (c)
200\(\Omega \)
- (d)
272.8\(\Omega \)
In the AC network shown in the figure, the phasor voltage VAB (in volt) is
- (a)
zero
- (b)
\(5\angle 30°\)
- (c)
\(12.5\angle 30°\)
- (d)
\(17\angle 30°\)
Consider the below figure:
For the circuit given above, the Thevenin's voltage across the terminals A and B is
- (a)
1.25 V
- (b)
0.25 V
- (c)
1 V
- (d)
0.5 V
Branch current and loop current relation is expressed in matrix form shown below, where I1 represents branch current and Ik represents loop current.
\(\left[ \begin{matrix} I_{ 1 } \\ I_{ 2 } \\ I_{ 3 } \\ I_{ 4 } \\ I_{ 5 } \\ I_{ 6 } \\ I_{ 7 } \\ I_{ 8 } \end{matrix} \right] =\left[ \begin{matrix} 0 \\ -1 \\ 0 \\ 1 \\ 0 \\ 1 \\ 1 \\ 0 \end{matrix}\begin{matrix} 0 \\ -1 \\ 1 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 1 \\ -1 \\ 0 \\ 0 \\ -1 \\ 0 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ -1 \\ -1 \\ 0 \\ 1 \end{matrix} \right] \left[ \begin{matrix} I_{ 1 } \\ I_{ 2 } \\ I_{ 3 } \\ I_{ 4 } \end{matrix} \right] \)
The directed graph will be
- (a)
- (b)
- (c)
- (d)
In the following circuit, the maximum power transfer condition is met for the load RL.
The value of RL will be
- (a)
2\(\Omega \)
- (b)
3\(\Omega \)
- (c)
1\(\Omega \)
- (d)
None of these