# 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
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$\cap$Q)=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.