Question 1 |

The value of the quantity P, where P=\int_{0}^{1}xe^{x}dx, is equal to

0 | |

1 | |

e | |

1/e |

Question 1 Explanation:

P=\int_{0}^{1}xe^x\; dx

Integrating by parts:

\begin{aligned} \text{Let } \; u&= x,\\ dv&= e^x\; dx\\ du&= dx,\\ v&=\int e^x\; dx=e^x \\ \text{Now, } \int udv&= uv-\int v\; du\\ \therefore \;\;\int x e^x\; dx &=xe^x-\int e^x\; dx \\ &=xe^x-e^x+c \\ \int_{0}^{1} xe^x\; dx&= [xe^x-e^x]_0^1\\ &= (1 \times e^1 -e^1)-(0 \times e^0 -e^0)\\ &= 0-(-1)\\ &= 1 \end{aligned}

Integrating by parts:

\begin{aligned} \text{Let } \; u&= x,\\ dv&= e^x\; dx\\ du&= dx,\\ v&=\int e^x\; dx=e^x \\ \text{Now, } \int udv&= uv-\int v\; du\\ \therefore \;\;\int x e^x\; dx &=xe^x-\int e^x\; dx \\ &=xe^x-e^x+c \\ \int_{0}^{1} xe^x\; dx&= [xe^x-e^x]_0^1\\ &= (1 \times e^1 -e^1)-(0 \times e^0 -e^0)\\ &= 0-(-1)\\ &= 1 \end{aligned}

Question 2 |

Divergence of the three-dimensional radial vector field \vec{r} is

3 | |

1/r | |

(\hat{i}+\hat{j}+\hat{k}) | |

3(\hat{i}+\hat{j}+\hat{k}) |

Question 2 Explanation:

\begin{aligned} \vec{r} &= x\hat{i}+y\hat{j}+z\hat{k}\\ \text{div } \vec{r}&= \bigtriangledown \cdot \vec{r}\\ &=\left ( \hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z} \right )(x\hat{i}+y\hat{j}+z\hat{k})\\ &=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}\\ &=1+1+1=3 \end{aligned}

Question 3 |

The period of the signal x(t)=8 sin(0.8 \pi t+ \pi/4) is

0.4 \pi s | |

0.8 \pi s | |

1.25 s | |

2.5 s |

Question 3 Explanation:

If period = T, then x(t+T)=x(t)

Here, x(t)= 8 \sin(0.8 \pi t+\pi /4)

putting x(t+T)=x(t) we get

8 \sin(0.8\pi (t+T)+\pi /4)=8 \sin (0.8 \pi t+\pi /4)

\sin(0.8\pi (t+T)+\pi /4)= \sin (0.8 \pi t+ \pi /4)

\sin(\theta _1)=\sin(\theta _2)\Rightarrow \theta _1=\theta _2+2 \pi

\therefore \;\; 0.8\pi (t+T)+ \pi /4=0.8 \pi t+ \pi /4+2 \pi

0.8\pi T=2 \pi

T=2.5 s

Here, x(t)= 8 \sin(0.8 \pi t+\pi /4)

putting x(t+T)=x(t) we get

8 \sin(0.8\pi (t+T)+\pi /4)=8 \sin (0.8 \pi t+\pi /4)

\sin(0.8\pi (t+T)+\pi /4)= \sin (0.8 \pi t+ \pi /4)

\sin(\theta _1)=\sin(\theta _2)\Rightarrow \theta _1=\theta _2+2 \pi

\therefore \;\; 0.8\pi (t+T)+ \pi /4=0.8 \pi t+ \pi /4+2 \pi

0.8\pi T=2 \pi

T=2.5 s

Question 4 |

The system represented by the input-output relationship y(t)=\int_{-\infty }^{5t}x(\tau )d\tau , \; t\gt0

Linear and causal | |

Linear but not causal | |

Causal but not linear | |

Neither liner nor causal |

Question 4 Explanation:

Integrator is always a linear system.

\begin{aligned} \text{Since, }y(t)&=\int_{-\infty }^{5t}x(\tau )d\tau \;\;t \gt 0 \\ &\text{For t=1,} \\ y(t)&= \int_{-\infty }^{5}x(\tau )d\tau \end{aligned}

here value at t=1 depends on future values like at t=2,3,... of input x(t).

So, it is a noncausal systems.

\begin{aligned} \text{Since, }y(t)&=\int_{-\infty }^{5t}x(\tau )d\tau \;\;t \gt 0 \\ &\text{For t=1,} \\ y(t)&= \int_{-\infty }^{5}x(\tau )d\tau \end{aligned}

here value at t=1 depends on future values like at t=2,3,... of input x(t).

So, it is a noncausal systems.

Question 5 |

The switch in the circuit has been closed for a long time. It is opened at t = 0. At t =0^+, the current through the 1 \muF capacitor is

0A | |

1A | |

1.25A | |

5A |

Question 5 Explanation:

As the switch has been closed for a long time, the circuit is in steady state. At steadystate, capacitor is open circuit,

Using KVL,

5-I-4I=0 I=1A

V_c(0^-)=4 \times 1=4V

As the voltage across capacitorcan not change abruptly,

So, V_c(0^+)=V_c(0^-)=4V

Circuit at t=0^+

Current through capacitor at t=0^+

I_c(0^+)=\frac{4}{4}=1A

Using KVL,

5-I-4I=0 I=1A

V_c(0^-)=4 \times 1=4V

As the voltage across capacitorcan not change abruptly,

So, V_c(0^+)=V_c(0^-)=4V

Circuit at t=0^+

Current through capacitor at t=0^+

I_c(0^+)=\frac{4}{4}=1A

Question 6 |

The second harmonic component of the periodic waveform given in the figure has an amplitude of

0 | |

1 | |

2/\pi | |

\sqrt{5} |

Question 6 Explanation:

The given signal is odd as wel as having half wave symmetry.

So, it has only sine terms with odd harmonics. So, for second harmonic term amplitude =0.

So, it has only sine terms with odd harmonics. So, for second harmonic term amplitude =0.

Question 7 |

As shown in the figure, a 1 \Omega resistance is connected across a source that has a load line v+i=100. The current through the resistance is

25A | |

50A | |

100A | |

200A |

Question 7 Explanation:

A resistor has linear characteristics

i.e. V=Ri

\Rightarrow \;\; V=i

Load line,

V+i=100

i+i=100

current through resistance,

i=\frac{100}{2}=50A

i.e. V=Ri

\Rightarrow \;\; V=i

Load line,

V+i=100

i+i=100

current through resistance,

i=\frac{100}{2}=50A

Question 8 |

A wattmeter is connected as shown in figure. The wattmeter reads.

Zero always | |

Total power consumed by Z_1 \; and \; Z_2 | |

Power consumed by Z_1 | |

Power consumed by Z_2 |

Question 8 Explanation:

Potential coil draws negligible current. So current through Z_1 and Z_2 is same.

current through current coil =I_{cc}=I

Voltage across potential coil =V_{pc}

Voltage across Z_1=V_{pc}=V

Wattmeter reads power consumed by Z_2, as voltage across potential coil = Voltage across Z_2

Current through current coil = Current through Z_2.

Question 9 |

An ammeter has a current range of 0-5 A, and its internal resistance is 0.2 \Omega. In
order to change the range to 0-25 A, we need to add a resistance of

0.8 \Omega in series with the meter | |

1.0 \Omega in series with the meter | |

0.04 \Omega in parallel with the meter | |

0.05 \Omega in parallel with the meter |

Question 9 Explanation:

To extend the range of ammeter, a resistance R_{sh} is connected across the meter.

I_m= full scale deflection current=5A

I=25A

Multiplying power =m=\frac{I}{I_m}=\frac{25}{5}=5

I_m=\frac{IR_{sh}}{R_{sh}+R_m}

\frac{I}{I_m}=\frac{R_{sh}+R_m}{R_{sh}}

m=1+\frac{R_m}{R_{sh}}

R_{sh}=\frac{R_m}{m-1}=\frac{0.2}{5-1}=0.05

I_m= full scale deflection current=5A

I=25A

Multiplying power =m=\frac{I}{I_m}=\frac{25}{5}=5

I_m=\frac{IR_{sh}}{R_{sh}+R_m}

\frac{I}{I_m}=\frac{R_{sh}+R_m}{R_{sh}}

m=1+\frac{R_m}{R_{sh}}

R_{sh}=\frac{R_m}{m-1}=\frac{0.2}{5-1}=0.05

Question 10 |

As shown in the figure, a negative feedback system has an amplifier of gain 100
with \pm10% tolerance in the forward path, and an attenuator of value 9/100 in
the feedback path. The overall system gain is approximately :

10\pm 1% | |

10\pm 2% | |

10\pm 5% | |

10\pm 10% |

Question 10 Explanation:

G=100\pm 10%

\frac{\Delta G}{G}=10% or 0.1

H=\frac{9}{100}

Overall gain,

T=\frac{G}{1+GH}

T=\frac{100}{1+100 \times \frac{9}{100}}=10

Sensitivity w.r.t. 'G'=\frac{1}{1+GH} \times 10%

\;\;\;=\frac{1}{1+10 \times 1}\times 10%

\;\;\;=\frac{10}{11}\cong 1%

\frac{\Delta G}{G}=10% or 0.1

H=\frac{9}{100}

Overall gain,

T=\frac{G}{1+GH}

T=\frac{100}{1+100 \times \frac{9}{100}}=10

Sensitivity w.r.t. 'G'=\frac{1}{1+GH} \times 10%

\;\;\;=\frac{1}{1+10 \times 1}\times 10%

\;\;\;=\frac{10}{11}\cong 1%

There are 10 questions to complete.