GATE EE 2005

Question 1
In the figure given below the value of R is
A
2.5\Omega
B
5.0\Omega
C
7.5\Omega
D
10.0\Omega
Electric Circuits   Basics
Question 1 Explanation: 
The resultant (R) when viewed from voltage source =\frac{100}{8}=12.5
R+10||10=12.5\Omega
\therefore \;\; R=12.5-10||10 =12.5-5=7.5\Omega
Question 2
The RMS value of the voltage u(t)= 3 + 4 cos (3t) is
A
\sqrt{17} V
B
5V
C
7V
D
(3+2\sqrt{2})V
Electric Circuits   Steady State AC Analysis
Question 2 Explanation: 
R.M.S. value of d.c voltage =V_{dc}^{(rms)}=3V
R.M.S. value of a.c. voltage =V_{ac}^{(rms)}=\left ( \frac{4}{\sqrt{2}} \right ) V
Therefore, R.M.S. value of the voltage
\sqrt{3^2+\left ( \frac{4}{\sqrt{2}} \right )^2}
=\sqrt{9+8}=\sqrt{17}V
Question 3
For the two port network shown in the figure, the Z-matrix is given by
A
\begin{bmatrix} Z_{1} &Z_{1}+Z_{2} \\ Z_{1}+Z_{2} & Z_{2} \end{bmatrix}
B
\begin{bmatrix} Z_{1} &Z_{1} \\ Z_{1}+Z_{2} & Z_{2} \end{bmatrix}
C
\begin{bmatrix} Z_{1} &Z_{2} \\ Z_{2} & Z_{1}+Z_{2} \end{bmatrix}
D
\begin{bmatrix} Z_{1} &Z_{1} \\ Z_{1} & Z_{1}+Z_{2} \end{bmatrix}
Electric Circuits   Two Port Network and Network Functions
Question 3 Explanation: 


v_1=(i_1+i_2)Z_1 =Z_1i_1+Z_1i_2
v_2=Z_2i_1+Z_1(i_1+i_2) =Z_1i_1+(Z_1+Z_2)i_2
From above, we can write,
\text{Z-matrix}=\begin{bmatrix} Z_1 &Z_1 \\ Z_1 & Z_1+Z_2 \end{bmatrix}
Question 4
In the figure given, the initial capacitor voltage is zero. The switch is closed at t = 0. The final steady-state voltage across the capacitor is
A
20V
B
10V
C
5V
D
0V
Electric Circuits   Transients and Steady State Response
Question 4 Explanation: 
At (t\rightarrow 0^+). The capacitor act as short-circuit. At (t\rightarrow \infty ), the capacitor will become open circuit.

Therefore, voltage across capacitor =\frac{20}{10+10}\times 10=10V
Question 5
If \vec{E} is the electric intensity, \triangledown \cdot (\triangledown \times \vec{E}) is equal to
A
\vec{E}
B
|\vec{E}|
C
null vector
D
Zero
Electromagnetic Fields   Coordinate Systems and Vector Calculus
Question 5 Explanation: 
Divergence of a curl field is always zero.
i.e. \triangledown \cdot (\triangledown \times \vec{E})=0
Question 6
A system with zero initial conditions has the closed loop transfer function T(s)=\frac{s^{2}+4}{(s+1)(s+4)}. The system output is zero at the frequency
A
0.5 rad/sec
B
1 rad/sec
C
2 rad/sec
D
4 rad/sec
Control Systems   Frequency Response Analysis
Question 6 Explanation: 
T(s)=\frac{s^2+4}{(s+1)(s+4)}
For frequency response, put s=j\omega
T(j\omega )=\frac{(j\omega )^2+4}{(j\omega +1)(j\omega +4)}=\frac{4-\omega ^2}{(1+j\omega )(4+j\omega )}
Magnitude of T(j\omega )=|T(j\omega )|=\frac{4-\omega ^2}{\sqrt{(1+\omega ^2)(16+\omega )^2}}=0
\omega =2 rad/sec
The system output is zero at 2 rad/sec.
Question 7
Figure shows the root locus plot (location of poles not given) of a third order system whose open loop transfer function is
A
\frac{K}{s^{3}}
B
\frac{K}{s^{2}(s+1)}
C
\frac{K}{s(s^{2}+1)}
D
\frac{K}{s(s^{2}-1)}
Control Systems   Root Locus Techniques
Question 7 Explanation: 


These are three asymptotes with angle 60^{\circ},180^{\circ} and 300^{\circ}
Angle of asymptotes =\frac{(2k+1)\times 180^{\circ}}{P-Z}
Where, k=0,1,2 up to (P-Z)-1 as angle are 60^{\circ},180^{\circ} and 300^{\circ}
it means P-Z=3
Intersection of asymptotes on real axis
x=\frac{\Sigma poles -\Sigma zeros}{P-Z}
Since, system does not have zeroes
x=\frac{\Sigma poles}{P}
As asymptotes intersect at origin, it means all the three poles are on right
Hence, Option (A) is correct.
Question 8
The gain margin of a unity feed back control system with the open loop transfer function G(s)=\frac{(s+1)}{s^{2}} is
A
0
B
\frac{1}{\sqrt{2}}
C
\sqrt{2}
D
\infty
Control Systems   Frequency Response Analysis
Question 8 Explanation: 


-1+j0 is not enclosed , system is always stable.
\therefore \;\; G.M.=\infty
Question 9
In the matrix equation Px=q, which of the following is a necessary condition for the existence of at least on solution for the unknown vector x:
A
Augmented matrix [Pq] must have the same rank as matrix P
B
Vector q must have only non-zero elements
C
Matrix P must be singular
D
Matrix P must be square
Engineering Mathematics   Linear Algebra
Question 9 Explanation: 
rank [Pq] = rank [P] is necessary for existence of at least one solution to Px=q.
Question 10
If P and Q are two random events, then the following is TRUE
A
Independence of P and Q implies that probability (P\capQ)=0
B
Probability (P \cup Q) \geq Probability (P) + Probability (Q)
C
If P and Q are mutually exclusive, then they must be independent
D
Probability (P \cap Q) \leq Probability (P)
Engineering Mathematics   Probability and Statistics
Question 10 Explanation: 
(A) is false since of P and Q are independent
Pr(P\cap Q)=Pr(P) \times Pr(Q)
which need not be zero.
(B) is false since
Pr(P\cup Q)=Pr(P) + Pr(Q)-Pr(P\cap Q)
\therefore \; Pr(P\cup Q)\leq Pr(P) + Pr(Q)
(C) is false since independence and mutually exclusion are unrelated properties.
(D) is true
\begin{aligned} \text{since } P\cap Q&\subseteq P \\ \Rightarrow \;\; n(P\cap Q) &\leq n(P) \\ \Rightarrow \;\; Pr(P\cap Q) &\leq Pr(P) \end{aligned}
There are 10 questions to complete.
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