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
H(s)=\frac{As+B}{s^2+Cs+D}
where A, B, C and D are positive constants. This system cannot operate as
low pass filter. | |
high pass filter | |
band pass filter. | |
an integrator. |
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.
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.)
12.14 | |
68.26 | |
51.1 | |
78.4 |
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
\frac{23s^{2}+46s+20}{4s+5} | |
6s+4 | |
7s+4 | |
\frac{25s^{2}+46s+20}{4s+5} |
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}
\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
0 A | |
\frac{100}{\sqrt{3}} A | |
100 A | |
300 A |
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
11 | |
12 | |
13 | |
14 |
Question 5 Explanation:
Loops =b-(N-1)
5=b-(8-1)
b=12
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
2 and 5 | |
5 and 2 | |
3 and 4 | |
4 and 3 |
Question 6 Explanation:
Number of KCL equations
= n-1=5-1=4
Number of KVL equations
=b-(n-1) = 7-(5-1)=3
= 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 __________.
100 | |
10 | |
20 | |
50 |
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.
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 ____.
0 | |
0.25 | |
0.5 | |
0.75 |
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
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= 0.5, b= 2.0 | |
a=2.0, b=0.5 | |
a=1.0, b=1.0 | |
a=4.0, b=0.5 |
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 _____.
10 | |
15 | |
55 | |
35 |
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
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.