Question 1 |

For matrices of same dimension M, N and scalar c, which one of these properties DOES NOT ALWAYS hold ?

(M^{T})^{T}=M | |

(cM)^{T}=c(M)^{T} | |

(M+N)^{T}=M^{T}+N^{T} | |

MN=NM |

Question 1 Explanation:

Matrix multiplication is not commutative.

Question 2 |

In a housing society, half of the families have a single child per family, while the
remaining half have two children per family. The probability that a child picked
at random, has a sibling is _____

1.5 | |

0.67 | |

2 | |

3 |

Question 2 Explanation:

Required probability

=\frac{\frac{1}{2} \times \frac{2}{3}}{\frac{1}{2} \times \frac{1}{3}+\frac{1}{2} \times \frac{2}{3}}=\frac{2}{3}=0.666

=\frac{\frac{1}{2} \times \frac{2}{3}}{\frac{1}{2} \times \frac{1}{3}+\frac{1}{2} \times \frac{2}{3}}=\frac{2}{3}=0.666

Question 3 |

C is a closed path in the z -plane by |z| = 3. The value of the integral \oint_{c}(\frac{z^{2}-z+4j}{z+2j})dz is

-4\pi (1+ j2 ) | |

4\pi (3- j2 ) | |

-4\pi (3+ j2 ) | |

4\pi (1- j2) |

Question 3 Explanation:

\oint_{c} \frac{z^{2}-z+4 j}{z+2 j}

Pole =z=-2 j

which is inside of |z|=3

From Cauchy integral formula

\begin{aligned} \oint \frac{z^{2}-z+4 j}{z+2 j} &=2 \pi i\left[\lim _{z \rightarrow-2 j} z^{2}-z+4 j\right] \\ &=2 \pi j[-4+2 j+4 j] \\ &=2 \pi j[-4+6 j] \\ &=-4 \pi[2 j+3] \end{aligned}

Question 4 |

A real (4x4) matrix A satisfies the equation A_{2} = I, where I is the (4x4)
identity matrix. The positive eigen value of A is _____.

1 | |

2 | |

3 | |

4 |

Question 4 Explanation:

\begin{aligned} \text{since, }A^{2}=l, \text{eig}\left(A^{2}\right)&=\text{eig}(I)=1 \\ \Rightarrow \quad \text{eig}(A)^{2}&=1\\ \Rightarrow \quad \text{eig}(A)&=\pm 1 \end{aligned}

Therefore, the positive eigen value of A is +1.

Therefore, the positive eigen value of A is +1.

Question 5 |

Let X_{1} , X_{2}, \; and \; X_{3} be independent and identically distributed random variables with the uniform distribution on [0,1]. The probability P{X_{1} is the largest} is

0.5 | |

0.33 | |

0.25 | |

0.75 |

Question 5 Explanation:

If multiple independent random variables are uniformly distributed in the same interval then each random variable will have equal chances to be largest and to be lowest.

P\left(X_{1} \text { is the largest) }=\frac{1}{3}\right.

P\left(X_{1} \text { is the largest) }=\frac{1}{3}\right.

Question 6 |

For maximum power transfer between two cascaded sections of an electrical
network, the relationship between the output impedance Z_{1} of the first section
to the input impedance Z_{2} of the second section is

Z_{2}=Z_{1} | |

Z_{2}=-Z_{1} | |

Z_{2}=Z_{1}^{*} | |

Z_{2}=-Z_{1}^{*} |

Question 7 |

Consider the configuration shown in the figure which is a portion of a larger
electrical network

For R = 1\Omega and currents i_{1} = 2 A , i_{4} =- 1A , i_{5}=- 4 A , which one of the following is TRUE ?

For R = 1\Omega and currents i_{1} = 2 A , i_{4} =- 1A , i_{5}=- 4 A , which one of the following is TRUE ?

i_{6}=5A | |

i_{3}=-4A | |

Data is sufficient to conclude that the supposed currents are impossible | |

Data is insufficient to identify the currents i_{2},i_{3} \; and \; i_{6} |

Question 7 Explanation:

Given data:

\begin{array}{l} i_{1}=2 \mathrm{A}, i_{4}=-1 \mathrm{A}, i_{5}=-4 \mathrm{A} \\ R=1 \Omega \end{array}

To calculate:

i_{6}=?

Using KVL at all the three nodes we get,

At node A

i_{5}-i_{3}+i_{2}=0\qquad \ldots(i)

At node B

i_{4}+i_{1}-i_{2}=0\qquad \ldots(ii)

At node C

i_{6}+i_{3}-i_{1}=0 \qquad \ldots(iii)

By putting the value of i_{3} and i_{2} from equation (i)

and (ii) in equation (iii) we get,

\begin{aligned} i_{6}+\left(i_{2}+i_{5}\right)-i_{1}&=0 \\ i_{6}+\left(i_{1}+i_{4}+i_{5}\right)-i_{1}&=0 \\ \therefore \quad i_{6}+(2-1-4)-2&=0 \\ i_{6}&=5 \mathrm{A} \end{aligned}

\begin{array}{l} i_{1}=2 \mathrm{A}, i_{4}=-1 \mathrm{A}, i_{5}=-4 \mathrm{A} \\ R=1 \Omega \end{array}

To calculate:

i_{6}=?

Using KVL at all the three nodes we get,

At node A

i_{5}-i_{3}+i_{2}=0\qquad \ldots(i)

At node B

i_{4}+i_{1}-i_{2}=0\qquad \ldots(ii)

At node C

i_{6}+i_{3}-i_{1}=0 \qquad \ldots(iii)

By putting the value of i_{3} and i_{2} from equation (i)

and (ii) in equation (iii) we get,

\begin{aligned} i_{6}+\left(i_{2}+i_{5}\right)-i_{1}&=0 \\ i_{6}+\left(i_{1}+i_{4}+i_{5}\right)-i_{1}&=0 \\ \therefore \quad i_{6}+(2-1-4)-2&=0 \\ i_{6}&=5 \mathrm{A} \end{aligned}

Question 8 |

When the optical power incident on a photodiode is 10\muW and the responsivity
is 0.8 A/W, the photocurrent generated (in \mu A) is _____.

2 | |

4 | |

8 | |

10 |

Question 8 Explanation:

\begin{array}{l} \text { Responsivity }(R)=\frac{I_{p}}{P_{o}} \\ \text { where } I_{p}=\text { Photo current } \\ \qquad P_{0}=\text { Incident power } \\ \therefore \quad I_{p}=R \times P_{0}=8 \mu \mathrm{A} \end{array}

Question 9 |

In the figure, assume that the forward voltage drops of the PN diode D_{1} and
Schottky diode D_{2} are 0.7 V and 0.3 V, respectively. If ON denotes conducting
state of the diode and OFF denotes non-conducting state of the diode, then in
the circuit,

both D_{1} \; and \; D_{2} are ON | |

D_{1} is ON and D_{2} are OFF | |

both D_{1} \; and \; D_{2} are OFF | |

D_{1} is OFF and D_{2} are ON |

Question 9 Explanation:

Consider D_{1} \rightarrow \text{OFF } and D_{2} \rightarrow \text{ON} then

Apply KVL

\begin{aligned} 10 &=1000 I+20 I+0.3 \\ I &=\frac{9.7}{1020} \\ I &=9.5 \mathrm{mA} \end{aligned}

Now, we calculate V_{D_{1}} of

\begin{aligned} & 10=9.5+V_{D_{1}} \\ \therefore V_{D_{1}} &=0.5 \mathrm{V} \end{aligned}

since v_{D_{1}} \lt 0.7 \mathrm{V}, \quad \mathrm{D}_{1} is in OFF state i.e. our assumption is correct and hence (D) is the correct option.

Apply KVL

\begin{aligned} 10 &=1000 I+20 I+0.3 \\ I &=\frac{9.7}{1020} \\ I &=9.5 \mathrm{mA} \end{aligned}

Now, we calculate V_{D_{1}} of

\begin{aligned} & 10=9.5+V_{D_{1}} \\ \therefore V_{D_{1}} &=0.5 \mathrm{V} \end{aligned}

since v_{D_{1}} \lt 0.7 \mathrm{V}, \quad \mathrm{D}_{1} is in OFF state i.e. our assumption is correct and hence (D) is the correct option.

Question 10 |

If fixed positive charges are present in the gate oxide of an n-channel
enhancement type MOSFET, it will lead to

a decrease in the threshold voltage | |

channel length modulation | |

an increase in substrate leakage current | |

an increase in accumulation capacitance |

Question 10 Explanation:

Fixed charges reduces threshold voltage.

There are 10 questions to complete.