Question 1 |

A benchtop dc power supply acts as an ideal 4 A current source as long as its terminal
voltage is below 10 V. Beyond this point, it begins to behave as an ideal 10 V voltage
source for all load currents going down to 0 A. When connected to an ideal rheostat,
find the load resistance value at which maximum power is transferred, and the corresponding
load voltage and current.

Short, \infty A, 10 V | |

Open, 4 A, 0 V | |

2.5 \Omega, 4 A, 10 V | |

2.5 \Omega, 4 A, 5 V |

Question 1 Explanation:

Maximum power transistor of VI product is maximum. If draw the the curve, it intersect (10, 4) that will give maximum power. The terminal voltage is 10 V (Load voltage) and current is 4 A (Load current). Load resistance is \frac{10}{4}=2.5\Omega .

Question 2 |

The Thevenin equivalent voltage, V_{TH}, in V (rounded off to 2 decimal places) of the network
shown below, is _______ .

5 | |

7 | |

14 | |

6 |

Question 2 Explanation:

Only voltage source 4V is there and current source 5A is open circuited

From the above circuit,

V_{TH1}=4V

Case-II:

Only current source 5A is there and voltage source 4V is short circuited.

From the above circuit,

V_{TH2}=2 \times 5=10V

By applying superposition theorem,

V_{TH}=V_{TH1}+V_{TH2}=10+4=14V

From the above circuit,

V_{TH1}=4V

Case-II:

Only current source 5A is there and voltage source 4V is short circuited.

From the above circuit,

V_{TH2}=2 \times 5=10V

By applying superposition theorem,

V_{TH}=V_{TH1}+V_{TH2}=10+4=14V

Question 3 |

x_R\; and \; x_A are, respectively, the rms and average values of x(t) = x(t - T), and similarly,
y_R\; and \; y_A are, respectively, the rms and average values of y(t) = kx(t). k, \;T are
independent of t. Which of the following is true?

y_A=kx_A;\; y_R=kx_R | |

y_A=kx_A;\; y_R\neq kx_R | |

y_A \neq kx_A;\; y_R= kx_R | |

y_A \neq kx_A;\; y_R\neq kx_R |

Question 3 Explanation:

Given that,

\begin{aligned} x(t)&=x(t-T) \; \text{ i.e. periodic signal}\\ \text{Average of } x(t)&=x_{A}\\ \text{Rms of }x(t)&=x_{R} \\ \text{Average of }y(t)&=y_{A}\\ \text{Rms of } y(t)&=y_{R} \\ y(t)&=k_{x}(t_{0}) \; \; ....(i) \\ &\text{Using equation(i),}\\ \text{Average of }y(t)&=k \times \text{ Average of }x(t) \\ y_{A}&=kx_{A}\\ \text{Power of }y(t)&=|k|^{2}\text{ Power of }x(t) \\ Rms^{2} \text{ of } y(t)&=|k|^{2}\; Rms^{2}\text{ of }x(t) \\ y_{R}^{2}=|k|^{2}\cdot x_{R}^{2} \\ y_{R}=|k|x_{R}\end{aligned}

\begin{aligned} x(t)&=x(t-T) \; \text{ i.e. periodic signal}\\ \text{Average of } x(t)&=x_{A}\\ \text{Rms of }x(t)&=x_{R} \\ \text{Average of }y(t)&=y_{A}\\ \text{Rms of } y(t)&=y_{R} \\ y(t)&=k_{x}(t_{0}) \; \; ....(i) \\ &\text{Using equation(i),}\\ \text{Average of }y(t)&=k \times \text{ Average of }x(t) \\ y_{A}&=kx_{A}\\ \text{Power of }y(t)&=|k|^{2}\text{ Power of }x(t) \\ Rms^{2} \text{ of } y(t)&=|k|^{2}\; Rms^{2}\text{ of }x(t) \\ y_{R}^{2}=|k|^{2}\cdot x_{R}^{2} \\ y_{R}=|k|x_{R}\end{aligned}

Question 4 |

The current flowing in the circuit shown below in amperes is _____

0 | |

1 | |

2 | |

4 |

Question 4 Explanation:

By Millman'e theorem,

E=\frac{\frac{200}{50}+\frac{160}{40}-\frac{100}{25}-\frac{80}{20}}{\frac{1}{50}+\frac{1}{40}+\frac{1}{25}+\frac{1}{20}}=0V

\frac{1}{R}=\frac{1}{50}+\frac{1}{40}+\frac{1}{25}+\frac{1}{20}

Simplified circuit,

\therefore \;\;I=0A

Question 5 |

A 0.1\mu F capacitor charged to 100 V is discharged through a 1 k\Omega resistor. The time in ms (round off to two decimal places) required for the voltage across the capacitor to drop to 1V is ______

0.25 | |

0.65 | |

0.45 | |

0.85 |

Question 5 Explanation:

\begin{aligned} v_c(t)&=V_0e^{-t/\tau } \\ V_0&=100V \\ \tau &=RC=(10^3)(10^{-7}) \\ &=10^{-4}sec \\ \therefore \;v_c(t)&=100e^{-10^4 t }V \end{aligned}

Let the time required by the voltage across the capacitor to drop to 1 V is t_1,

\begin{aligned} \therefore \; v_c(t_1)&=100e^{-10^4t_1} \\ \text{But, } v_c(t_1)&=0 \\ \text{So, }0&=100e^{-10^4t_1} \\ t_1&=0.46msec \end{aligned}

Question 6 |

The line currents of a three-phase four wire system are square waves with amplitude of 100 A. These three currents are phase shifted by 120^{\circ} with respect to each other. The rms value of neutral current is

0 A | |

\frac{100}{\sqrt{3}} A | |

100 A | |

300 A |

Question 6 Explanation:

I_N=I_a+I_b+I_c

(I_N)_{rms}=100A

Question 7 |

The current I flowing in the circuit shown below in amperes (round off to one decimal place) is ____

0.8 | |

1.1 | |

1.4 | |

1.9 |

Question 7 Explanation:

Applying nodal at node x,

-I-2+\frac{V_x-5I}{3}=0

-3I-6+V_x-5I=0

\Rightarrow \; 8I=V_x-6\;\;...(i)

As, I=\frac{20-V_x}{2}

\Rightarrow \; V_x=20-2I\;\;...(ii)

Substituting (ii) in (i),

8I=20-2I-6

10I=14

I=1.4A

Question 8 |

The voltage v(t) across the terminals a and b as shown in the figure, is a sinusoidal voltage
having a frequency \omega=100 radian/s. When the inductor current i(t) is in phase with the
voltage v(t), the magnitude of the impedance Z (in \Omega) seen between the terminals a and b
is ________ (up to 2 decimal places).

25 | |

50 | |

100 | |

150 |

Question 8 Explanation:

At resonance imaginary part of Z_{eq}=0

Real of Z_{eq}=\frac{R_1 X_c^2}{R_1^2+X-c^2}

\;\;=\frac{100 \times 100 \times 100}{100^2+100^2}=50\Omega

Real of Z_{eq}=\frac{R_1 X_c^2}{R_1^2+X-c^2}

\;\;=\frac{100 \times 100 \times 100}{100^2+100^2}=50\Omega

Question 9 |

The voltage across the circuit in the figure, and the current through it, are given by the
following expressions:

v(t) = 5 - 10 cos(\omega t + 60^{\circ}) V

i(t) = 5 + X cos(\omega t) A

where \omega =100 \pi radian/s. If the average power delivered to the circuit is zero, then the value of X (in Ampere) is _____ (up to 2 decimal places).

v(t) = 5 - 10 cos(\omega t + 60^{\circ}) V

i(t) = 5 + X cos(\omega t) A

where \omega =100 \pi radian/s. If the average power delivered to the circuit is zero, then the value of X (in Ampere) is _____ (up to 2 decimal places).

5 | |

10 | |

15 | |

20 |

Question 9 Explanation:

Given that,

v(t)=5-10 \cos (\omega t+60^{\circ})

i(t)=5+C \cos (\omega t-0^{\circ})

P_{req}=0

0=5 \times 5 +\frac{1}{2}[(-10)(X)\cos(60^{\circ})]

-25=\frac{1}{2}[(-10)(X)\cos(60^{\circ})]

X=10

v(t)=5-10 \cos (\omega t+60^{\circ})

i(t)=5+C \cos (\omega t-0^{\circ})

P_{req}=0

0=5 \times 5 +\frac{1}{2}[(-10)(X)\cos(60^{\circ})]

-25=\frac{1}{2}[(-10)(X)\cos(60^{\circ})]

X=10

Question 10 |

The equivalent impedance Z_{eq} for the infinite ladder circuit shown in the figure is

j12 \Omega | |

-j12 \Omega | |

j13 \Omega | |

13 \Omega |

Question 10 Explanation:

Z_1=j9

Z_2=j5-j1=j4

Z_{eq}=Z_1+\frac{Z_2Z_{eq}}{Z_2+Z_{eq}}

By solving above equation,

Z_{eq}=j12

There are 10 questions to complete.