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.