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.