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


There are 5 questions to complete.