Question 1 |

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 1 Explanation:

I_N=I_a+I_b+I_c

(I_N)_{rms}=100A

Question 2 |

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 2 Explanation:

Loops =b-(N-1)

5=b-(8-1)

b=12

5=b-(8-1)

b=12

Question 3 |

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 3 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 4 |

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 4 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 5 |

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 5 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 6 |

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 6 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 7 |

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 7 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

Question 8 |

The following arrangement consists of an ideal transformer and an attenuator
which attenuates by a factor of 0.8. An ac voltage V_{WX1} = 100 V is applied
across WX to get an open circuit voltage V_{YZ1} across YZ. Next, an ac voltageV_{YZ2} = 100 V is applied across YZ to get an open circuit voltage V_{WX2} across WX. Then,V_{YZ1}/V_{WX1},V_{WX2}/V_{YZ2} are respectively,

125/100 and 80/100 | |

100/100 and 80/100 | |

100/100 and 100/100 | |

80/100 and 80/100 |

Question 8 Explanation:

V_{YZ_1}=100\times 1.25 \times 0.8 =100

In second case when 100 V is applied at YZ terminals, this whole 100 V will appear across the secondary winding,

Hence, V_{WX_2}=\frac{100}{1.25}=80V

\Rightarrow \; \frac{V_{YZ_1}}{V_{WX_1}}=\frac{100}{100}

\frac{V_{WX_2}}{V_{YZ_2}}=\frac{80}{100}

In second case when 100 V is applied at YZ terminals, this whole 100 V will appear across the secondary winding,

Hence, V_{WX_2}=\frac{100}{1.25}=80V

\Rightarrow \; \frac{V_{YZ_1}}{V_{WX_1}}=\frac{100}{100}

\frac{V_{WX_2}}{V_{YZ_2}}=\frac{80}{100}

Question 9 |

The number of chords in the graph of the given circuit will be

3 | |

4 | |

5 | |

6 |

Question 9 Explanation:

Number of branches =b =6

No. of nodes = n= 4

No. of chords =b-(n-1) =6-(4-1)=3

Question 10 |

The matrix A given below in the node incidence matrix of a network. The columns
correspond to branches of the network while the rows correspond to nodes. Let
V=[v_{1} \; v_{2} ... v_{6}]^{T} denote the vector of branch voltages while I=[i_{1} \; i_{2} ... i_{6}]^{T} that of branch currents. The vector E=[e_{1} \; e_{2} \; e_{3} \; e_{4}]^{T} denotes the vector of node voltages relative to a common ground.

\begin{bmatrix} 1 & 1&1 & 0&0 & 0\\ 0& -1&0 & -1& 1& 0\\ -1 &0 & 0 & 0 &-1 & -1\\ 0&0 & -1&1 & 0 &1 \end{bmatrix}

Which of the following statement is true ?

\begin{bmatrix} 1 & 1&1 & 0&0 & 0\\ 0& -1&0 & -1& 1& 0\\ -1 &0 & 0 & 0 &-1 & -1\\ 0&0 & -1&1 & 0 &1 \end{bmatrix}

Which of the following statement is true ?

The equations v_{1}-v_{2}+v_{3}=0,v_{3}+v_{4}-v_{5}=0 are KVL equations for the network for some loops | |

The equations v_{1}-v_{3}-v_{6}=0,v_{4}+v_{5}-v_{6}=0 are KVL equations for the network for some loops | |

E = AV | |

AV = 0 are KVI equations for the network |

Question 10 Explanation:

Convention:

(when branch b_j leaves away from node N_i )

(when branch b_j enters node N_i)

Given node incidence matrix

eg.: Branch b_1 enter N_3 and leaves N_1

The oriented network graph

Using KVL, in loop containing branches b_1,b_3 and b_6

v_1-v_6-v_3=0

In loop containing branches b_4,b_5 and b_6

v_4+v_5-v_6=0

(when branch b_j leaves away from node N_i )

(when branch b_j enters node N_i)

Given node incidence matrix

eg.: Branch b_1 enter N_3 and leaves N_1

The oriented network graph

Using KVL, in loop containing branches b_1,b_3 and b_6

v_1-v_6-v_3=0

In loop containing branches b_4,b_5 and b_6

v_4+v_5-v_6=0

There are 10 questions to complete.