Question 1 |

The pressure coil of a dynamometer type wattmeter is

Highly inductive | |

Highly resistive | |

Purely resistive | |

Purely inductive |

Question 1 Explanation:

It is difficult to have purely resistive pressure coil. The pressure coil has a small value of inductance, due to which error occurs in wattmeter readings.

Question 2 |

The measurement system shown in the figure uses three sub-systems in cascade
whose gains are specified as G_1,G_2 \; and \; \frac{1}{G_3}. The relative small errors associated with
each respective subsystem G_1,G_2 \; and \; G_3 are \varepsilon _1,\varepsilon _2 \; and \; \varepsilon _3. The error associated
with the output is :

\varepsilon _{1}+ \varepsilon _{2} +1/ \varepsilon _{3} | |

\frac{\varepsilon _{1}\varepsilon _{2}} {\varepsilon _{3}} | |

\varepsilon _{1}+ \varepsilon _{2} -\varepsilon _{3} | |

\varepsilon _{1}+ \varepsilon _{2} +\varepsilon _{3} |

Question 2 Explanation:

\frac{dG_1}{G_1}=\varepsilon _1, \frac{dG_2}{G_2}=\varepsilon _2,and \frac{dG_3}{G_3}=\varepsilon _3

Output (y)_0=\frac{G_1G_3}{G_3}x

where, \;\; x=input

\ln y=\ln G_1+ \ln G_2- \ln G_3 +\ln x

Differentiating both side,

\frac{dy}{y}=\frac{dG_1}{G_1}+\frac{dG_2}{G_2}-\frac{dG_3}{G_3}+\frac{dx}{x}

No error is specified in input, so \frac{dx}{x}=0

\frac{dy}{y}==\varepsilon _1+\varepsilon _2-\varepsilon _3

Output (y)_0=\frac{G_1G_3}{G_3}x

where, \;\; x=input

\ln y=\ln G_1+ \ln G_2- \ln G_3 +\ln x

Differentiating both side,

\frac{dy}{y}=\frac{dG_1}{G_1}+\frac{dG_2}{G_2}-\frac{dG_3}{G_3}+\frac{dx}{x}

No error is specified in input, so \frac{dx}{x}=0

\frac{dy}{y}==\varepsilon _1+\varepsilon _2-\varepsilon _3

Question 3 |

The following circuit has a source voltage V_s as shown in the graph. The current
through the circuit is also shown.

The element connected between a and b could be

The element connected between a and b could be

A | |

B | |

C | |

D |

Question 3 Explanation:

Diode act as a switch. When forward biased it is short circuited. But when suddenly reverse biased, current does not become zero instantly , initially the same current flow in opposite direction and after some time (turn off time) it will become zero.

Question 4 |

The two inputs of a CRO are fed with two stationary periodic signals. In the X-Y
mode, the screen shows a figure which changes from ellipse to circle and back
to ellipse with its major axis changing orientation slowly and repeatedly. The
following inference can be made from this.

The signals are not sinusoidal | |

The amplitudes of the signals are very close but not equal | |

The signals are sinusoidal with their frequencies very close but not equal | |

There is a constant but small phase difference between the signals |

Question 4 Explanation:

Because phase difference only patterns changes from ellipse to circle and back to ellipse.

Question 5 |

The increasing order of speed of data access for the following device is

(I) Cache Memory

(II) CD-ROM

(III) Dynamic RAM

(IV) Processor Registers

(V) Magnetic Tape

(I) Cache Memory

(II) CD-ROM

(III) Dynamic RAM

(IV) Processor Registers

(V) Magnetic Tape

(V), (II), (III), (IV), (I) | |

(V), (II), (III), (I), (IV) | |

(II), (I), (III), (IV), (V) | |

(V), (II), (I), (III), (IV) |

Question 5 Explanation:

Access time of the residter is very less than that from a memory access. So speed of data access is faster in case of processor register.

Second highest is cache memory because it's size is small, so searching of data takes less time.

Second highest is cache memory because it's size is small, so searching of data takes less time.

Question 6 |

A field excitation of 20 A in a certain alternator results in an armature current
of 400 A in short circuit and a terminal voltage of 2000 V on open circuit. The
magnitude of the internal voltage drop within the machine at a load current of 200 A is

1V | |

10V | |

100V | |

1000V |

Question 6 Explanation:

If R_a is neglected,

\vec{E_f}=\vec{V_t}+j\vec{I_a}X_s

For a given field current under short-circuit condition,

\begin{aligned} I_a &=I_{sc} \\ V_t&= 0\\ X_s &=\frac{E_f}{I_{sc}} \\ \text{But,}\;\; E_f &=V_{oc} \end{aligned}

(open circuit voltage i.e. I_a=0 with the same field current).

Then with the linearity assumption

\begin{aligned} X_s&=\left.\begin{matrix} \frac{V_{oc}}{I_{sc}} \end{matrix}\right|_{I_f=\text{constant}} \\ &= \frac{2000}{400}=5\Omega \end{aligned}

Internal voltage drop

I_aX_s=200 \times 5=1000V

\vec{E_f}=\vec{V_t}+j\vec{I_a}X_s

For a given field current under short-circuit condition,

\begin{aligned} I_a &=I_{sc} \\ V_t&= 0\\ X_s &=\frac{E_f}{I_{sc}} \\ \text{But,}\;\; E_f &=V_{oc} \end{aligned}

(open circuit voltage i.e. I_a=0 with the same field current).

Then with the linearity assumption

\begin{aligned} X_s&=\left.\begin{matrix} \frac{V_{oc}}{I_{sc}} \end{matrix}\right|_{I_f=\text{constant}} \\ &= \frac{2000}{400}=5\Omega \end{aligned}

Internal voltage drop

I_aX_s=200 \times 5=1000V

Question 7 |

The current through the 2 k\Omega resistance in the circuit shown is

0 mA | |

1 mA | |

2 mA | |

6 mA |

Question 7 Explanation:

Bridge is balanced i.i. node C and node D are at same potential. Therefore, no current flows through 2 k\Omega resistor.

Question 8 |

Out of the following plant categories

(i) Nuclear

(ii) Run-of-river

(iii) Pump Storage

(iv) Diesel

The base load power plant are

(i) Nuclear

(ii) Run-of-river

(iii) Pump Storage

(iv) Diesel

The base load power plant are

(i) and (ii) | |

(ii) and (iii) | |

(i), (ii) and (iv) | |

(i), (iii) and (iv) |

Question 8 Explanation:

Pumped storage plants and diesel stations supply power during peak loads.

NOTE: Base load-plant

(i) Low operating cost

(ii) Capability of working continuously long period.

NOTE: Base load-plant

(i) Low operating cost

(ii) Capability of working continuously long period.

Question 9 |

For a fixed value of complex power flow in a transmission line having a sending
end voltage V, the real loss will be proportional to

V | |

V^{2} | |

1/V^{2} | |

1/V |

Question 9 Explanation:

S= Complex power

S is defined as, S=VI

I=\frac{S}{V}

If R is the resistance of transmission line

Real power loss =I^2R=\left ( \frac{S}{V} \right )^2R=\frac{S^2R}{V^2}

Since, S and R are fixed

Real power loss \propto \frac{1}{V^2}

S is defined as, S=VI

I=\frac{S}{V}

If R is the resistance of transmission line

Real power loss =I^2R=\left ( \frac{S}{V} \right )^2R=\frac{S^2R}{V^2}

Since, S and R are fixed

Real power loss \propto \frac{1}{V^2}

Question 10 |

How many 200 W/220 V incandescent lamps connected in series would consume
the same total power as a single 100 W/220 V incandescent lamp ?

not possible | |

4 | |

3 | |

2 |

Question 10 Explanation:

Let resistance of a single incandescent lamp=R.

Power consumed by a single lamp, P=200W.

When connected across voltage, V=220V.

So, P=\frac{V^2}{R}\; \Rightarrow \; 200=\frac{220^2}{R}

\Rightarrow \; R=242\Omega

Let, n number of lamps are connected in series across voltage V=200V.

So total resistance of lamps

R_{eq.}=nR=242n

Total power consumed,

P=\frac{V^2}{R_{eq.}}

\Rightarrow \; 100=\frac{220^2}{242n} \Rightarrow \; n=2

Power consumed by a single lamp, P=200W.

When connected across voltage, V=220V.

So, P=\frac{V^2}{R}\; \Rightarrow \; 200=\frac{220^2}{R}

\Rightarrow \; R=242\Omega

Let, n number of lamps are connected in series across voltage V=200V.

So total resistance of lamps

R_{eq.}=nR=242n

Total power consumed,

P=\frac{V^2}{R_{eq.}}

\Rightarrow \; 100=\frac{220^2}{242n} \Rightarrow \; n=2

There are 10 questions to complete.