# 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 $\alpha$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 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 $\mu$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 10 Explanation:
The two possible series connection are shown below:
Let the mutual inductance be M

$L_{eqv}=L_1+L_2-2M=240\mu H$
Thus, $L_1+L_2+2M=380 \mu H\;\;...(i)$
and $L_1+L_2-2M=240 \mu H\;\;...(ii)$
$4M=140\mu H$
$M=35\mu H$
Therefore, mutual inductance $M=35\mu H$