Magnetically Coupled Circuits, Network Topology and Filters

Question 1
The transfer function of a real system, H(s), is given as:
H(s)=\frac{As+B}{s^2+Cs+D}
where A, B, C and D are positive constants. This system cannot operate as
A
low pass filter.
B
high pass filter
C
band pass filter.
D
an integrator.
GATE EE 2022   Electric Circuits
Question 1 Explanation: 
Put s=0, H(0)=\frac{A \times 0+B}{0+C \times 0+D}=\frac{B}{D}
So, the system pass low frequency component. Put s=\infty , H(\infty )=0
For high pass filter, high frequency component should be non zero. Hence this system cannot be operated as high pass filter.
Question 2
An air-core radio-frequency transformer as shown has a primary winding and a secondary winding. The mutual inductance M between the windings of the transformer is ____________ \mu H.(Round off to 2 decimal places.)

A
12.14
B
68.26
C
51.1
D
78.4
GATE EE 2021   Electric Circuits
Question 2 Explanation: 


\begin{aligned} I_{1}&=\frac{5}{22}(p-p) \\ V_{0}&=j \omega M I_{1}=7.3=\left(2 \pi \times 10 \times 10^{3}\right) \times M \times\left(\frac{5}{22}\right) \\ M&=51.10 \mu \mathrm{H} \end{aligned}
Question 3
The input impedance, Z_{in}\left ( s \right ) for the network shown is

A
\frac{23s^{2}+46s+20}{4s+5}
B
6s+4
C
7s+4
D
\frac{25s^{2}+46s+20}{4s+5}
GATE EE 2021   Electric Circuits
Question 3 Explanation: 
Circuit in s-domain,

\begin{aligned} -S I_{1}+(4 s+5) I_{2} &=0 \\ \Rightarrow\qquad \qquad I_{2} &=\frac{s}{4 s+5} I_{1} \\ V_{1} &=(4+6 s) I_{1}-\frac{s^{2}}{4 s+5} I_{1} \\ \frac{V_{1}}{I_{1}} &=\frac{(4+6 s)(4 s+5)-s^{2}}{4 s+5} \\ &=\frac{24 s^{2}+30 s+16 s+20-s^{2}}{4 s+5} \\ Z_{\text {in }} &=\frac{23 s^{2}+46 s+20}{4 s+5} \end{aligned}
Question 4
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 4 Explanation: 


I_N=I_a+I_b+I_c
(I_N)_{rms}=100A
Question 5
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 5 Explanation: 
Loops =b-(N-1)
5=b-(8-1)
b=12
Question 6
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 6 Explanation: 
Number of KCL equations
= n-1=5-1=4
Number of KVL equations
=b-(n-1) = 7-(5-1)=3
Question 7
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 7 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 8
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 8 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 9
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 9 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 10
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 10 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
There are 10 questions to complete.