Magnetically Coupled Circuits, Network Topology and Filters

Question 1
The line currents of a three-phase four wire system are square waves with amplitude of 100 A. These three currents are phase shifted by 120^{\circ} with respect to each other. The rms value of neutral current is
A
0 A
B
\frac{100}{\sqrt{3}} A
C
100 A
D
300 A
GATE EE 2019   Electric Circuits
Question 1 Explanation: 


I_N=I_a+I_b+I_c
(I_N)_{rms}=100A
Question 2
The graph of a network has 8 nodes and 5 independent loops. The number of branches of the graph is
A
11
B
12
C
13
D
14
GATE EE 2018   Electric Circuits
Question 2 Explanation: 
Loops =b-(N-1)
5=b-(8-1)
b=12
Question 3
The graph associated with an electrical network has 7 branches and 5 nodes. The number of independent KCL equations and the number of independent KVL equations, respectively, are
A
2 and 5
B
5 and 2
C
3 and 4
D
4 and 3
GATE EE 2016-SET-2   Electric Circuits
Question 3 Explanation: 
Number of KCL equations
= n-1=5-1=4
Number of KVL equations
=b-(n-1) = 7-(5-1)=3
Question 4
The following figure shows the connection of an ideal transformer with primary to secondary turns ratio of 1:100. The applied primary voltage is 100 V (rms), 50 Hz, AC. The rms value of the current I, in ampere, is __________.
A
100
B
10
C
20
D
50
GATE EE 2016-SET-2   Electric Circuits
Question 4 Explanation: 
The above circuit can be drown by transferring secondary circuit to primary side.

I=\frac{100V}{(8+10j-4j)\Omega }
\;\;=\frac{100V}{(8+6j)\Omega }
So the rms value of I will be 10 A.
Question 5
Two identical coils each having inductance L are placed together on the same core. If an overall inductance of \alphaL is obtained by interconnecting these two coils, the minimum value of \alpha is ____.
A
0
B
0.25
C
0.5
D
0.75
GATE EE 2015-SET-2   Electric Circuits
Question 5 Explanation: 
CASE-I:
L_{eff}=L_+L_2=2L,, a=2
CASE-II:
l_{eff}=\frac{L_1L_2}{L_1+L_2}=\frac{L^2}{2L}=\frac{L}{2}, a=0.5
CASE-III
If both are diffferentially coupled then
L_{eff}=0
Minimum value =0
Question 6
Find the transformer ratios a and b such that the impedance (Z_{in}) is resistive and equals 2.5\Omega when the network is excited with a sine wave voltage of angular frequency of 5000 rad/s.
A
a= 0.5, b= 2.0
B
a=2.0, b=0.5
C
a=1.0, b=1.0
D
a=4.0, b=0.5
GATE EE 2015-SET-2   Electric Circuits
Question 6 Explanation: 


(R_{eq})_B=\frac{2.5}{a^2}
(2eq)_B=\left [ \frac{2.5}{a^2}+j\omega (1 \times 10^{-3})\right ]
(2eq)_A={\left [ \frac{2.5}{a^2}+j\omega (1 \times 10^{-3})\right ]}/b^2
Z_{in}=\frac{2.5}{a^2b^2}+\frac{ j\omega 10^{-3}}{b^2}+\frac{1}{j\omega (10 \times 10^{-5})}
Z_{in}=\frac{2.5}{a^2b^2}+j\left [ \frac{5 \times 10^3 \times 10^{-3}}{b^2} -\frac{1}{5 \times 10^3 \times 10 \times10^{-6}}\right ] \;...(i)
From problem,
Z_{in}=2.5 \;\;...(ii)
From equation (i) and (ii),
a^2b^2=1
\frac{5}{b^2}-\frac{1}{5 \times 10^{-2}}=0
\Rightarrow \; b=0.5 \text{ and } a=2
Question 7
Two identical coupled inductors are connected in series. The measured inductances for the two possible series connections are 380 \muH and 240 \muH. Their mutual inductance in \muH is _____.
A
10
B
15
C
55
D
35
GATE EE 2014-SET-2   Electric Circuits
Question 7 Explanation: 
The two possible series connection are shown below:
Let the mutual inductance be M

(i) Additive connection,

(ii) Substractive connection,
L_{eqv}=L_1+L_2-2M=240\mu H
Thus, L_1+L_2+2M=380 \mu H\;\;...(i)
and L_1+L_2-2M=240 \mu H\;\;...(ii)
Solving equations (i) and (ii), we get:
4M=140\mu H
M=35\mu H
Therefore, mutual inductance M=35\mu H
Question 8
The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage V_{WX1} = 100 V is applied across WX to get an open circuit voltage V_{YZ1} across YZ. Next, an ac voltageV_{YZ2} = 100 V is applied across YZ to get an open circuit voltage V_{WX2} across WX. Then,V_{YZ1}/V_{WX1},V_{WX2}/V_{YZ2} are respectively,
A
125/100 and 80/100
B
100/100 and 80/100
C
100/100 and 100/100
D
80/100 and 80/100
GATE EE 2013   Electric Circuits
Question 8 Explanation: 
V_{YZ_1}=100\times 1.25 \times 0.8 =100
In second case when 100 V is applied at YZ terminals, this whole 100 V will appear across the secondary winding,
Hence, V_{WX_2}=\frac{100}{1.25}=80V
\Rightarrow \; \frac{V_{YZ_1}}{V_{WX_1}}=\frac{100}{100}
\frac{V_{WX_2}}{V_{YZ_2}}=\frac{80}{100}
Question 9
The number of chords in the graph of the given circuit will be
A
3
B
4
C
5
D
6
GATE EE 2008   Electric Circuits
Question 9 Explanation: 


Number of branches =b =6
No. of nodes = n= 4
No. of chords =b-(n-1) =6-(4-1)=3
Question 10
The matrix A given below in the node incidence matrix of a network. The columns correspond to branches of the network while the rows correspond to nodes. Let V=[v_{1} \; v_{2} ... v_{6}]^{T} denote the vector of branch voltages while I=[i_{1} \; i_{2} ... i_{6}]^{T} that of branch currents. The vector E=[e_{1} \; e_{2} \; e_{3} \; e_{4}]^{T} denotes the vector of node voltages relative to a common ground.
\begin{bmatrix} 1 & 1&1 & 0&0 & 0\\ 0& -1&0 & -1& 1& 0\\ -1 &0 & 0 & 0 &-1 & -1\\ 0&0 & -1&1 & 0 &1 \end{bmatrix}

Which of the following statement is true ?
A
The equations v_{1}-v_{2}+v_{3}=0,v_{3}+v_{4}-v_{5}=0 are KVL equations for the network for some loops
B
The equations v_{1}-v_{3}-v_{6}=0,v_{4}+v_{5}-v_{6}=0 are KVL equations for the network for some loops
C
E = AV
D
AV = 0 are KVI equations for the network
GATE EE 2007   Electric Circuits
Question 10 Explanation: 
Convention:

(when branch b_j leaves away from node N_i )

(when branch b_j enters node N_i)
Given node incidence matrix

eg.: Branch b_1 enter N_3 and leaves N_1
The oriented network graph

Using KVL, in loop containing branches b_1,b_3 and b_6
v_1-v_6-v_3=0
In loop containing branches b_4,b_5 and b_6
v_4+v_5-v_6=0
There are 10 questions to complete.