# 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 $\alpha$L 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 $\mu$H and 240 $\mu$H. Their mutual inductance in $\mu$H 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  (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 voltage$V_{YZ2}$ = 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. 