Question 1 |
In the figure given below the value of R is
2.5\Omega | |
5.0\Omega | |
7.5\Omega | |
10.0\Omega |
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
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
\sqrt{17} V | |
5V | |
7V | |
(3+2\sqrt{2})V |
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
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
\begin{bmatrix} Z_{1} &Z_{1}+Z_{2} \\ Z_{1}+Z_{2} & Z_{2} \end{bmatrix} | |
\begin{bmatrix} Z_{1} &Z_{1} \\ Z_{1}+Z_{2} & Z_{2} \end{bmatrix} | |
\begin{bmatrix} Z_{1} &Z_{2} \\ Z_{2} & Z_{1}+Z_{2} \end{bmatrix} | |
\begin{bmatrix} Z_{1} &Z_{1} \\ Z_{1} & Z_{1}+Z_{2} \end{bmatrix} |
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
20V | |
10V | |
5V | |
0V |
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
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
\vec{E} | |
|\vec{E}| | |
null vector | |
Zero |
Question 5 Explanation:
Divergence of a curl field is always zero.
i.e. \triangledown \cdot (\triangledown \times \vec{E})=0
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
0.5 rad/sec | |
1 rad/sec | |
2 rad/sec | |
4 rad/sec |
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.
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
\frac{K}{s^{3}} | |
\frac{K}{s^{2}(s+1)} | |
\frac{K}{s(s^{2}+1)} | |
\frac{K}{s(s^{2}-1)} |
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
0 | |
\frac{1}{\sqrt{2}} | |
\sqrt{2} | |
\infty |
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:
Augmented matrix [Pq] must have the same rank as matrix P | |
Vector q must have only non-zero elements | |
Matrix P must be singular | |
Matrix P must be square |
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
Independence of P and Q implies that probability (P\capQ)=0 | |
Probability (P \cup Q) \geq Probability (P) + Probability (Q) | |
If P and Q are mutually exclusive, then they must be independent | |
Probability (P \cap Q) \leq Probability (P) |
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}
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.