GATE EE 2009

Question 1
The pressure coil of a dynamometer type wattmeter is
A
Highly inductive
B
Highly resistive
C
Purely resistive
D
Purely inductive
Electrical and Electronic Measurements   Measurement of Energy and Power
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 :
A
\varepsilon _{1}+ \varepsilon _{2} +1/ \varepsilon _{3}
B
\frac{\varepsilon _{1}\varepsilon _{2}} {\varepsilon _{3}}
C
\varepsilon _{1}+ \varepsilon _{2} -\varepsilon _{3}
D
\varepsilon _{1}+ \varepsilon _{2} +\varepsilon _{3}
Electrical and Electronic Measurements   Characteristics of Instruments and Measurement Systems
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
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
A
A
B
B
C
C
D
D
Analog Electronics   Miscellaneous
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.
A
The signals are not sinusoidal
B
The amplitudes of the signals are very close but not equal
C
The signals are sinusoidal with their frequencies very close but not equal
D
There is a constant but small phase difference between the signals
Electrical and Electronic Measurements   CRO and Electronic Measurements
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
A
(V), (II), (III), (IV), (I)
B
(V), (II), (III), (I), (IV)
C
(II), (I), (III), (IV), (V)
D
(V), (II), (I), (III), (IV)
Digital Electronics   Microprocessors
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.
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
A
1V
B
10V
C
100V
D
1000V
Electrical Machines   Synchronous Machines
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
Question 7
The current through the 2 k\Omega resistance in the circuit shown is
A
0 mA
B
1 mA
C
2 mA
D
6 mA
Electric Circuits   Basics
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
A
(i) and (ii)
B
(ii) and (iii)
C
(i), (ii) and (iv)
D
(i), (iii) and (iv)
Power Systems   Generating Power Stations
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.
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
A
V
B
V^{2}
C
1/V^{2}
D
1/V
Power Systems   Performance of Transmission Lines, Line Parameters and Corona
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}
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 ?
A
not possible
B
4
C
3
D
2
Electric Circuits   Basics
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
There are 10 questions to complete.
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