Electronic Science - Online Test

UGC NET Entrance exams Science Electronic Science TEST : 17

UGC NET Entrance exams Science # Electronic Science Electric Circuits Prepare / Learn

Q1.

With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed:

(i) 1Ω connected at port B draws a current of 3 A

(ii) 2.5Ω W connected at port B draws a current of 2 A

For the same network, with 6 V dc connected at port A, 1Ω connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is

A. 6 V

B. 7 V

C. 8 V

D. 9V

Answer : Option B
Explaination / Solution:
No Explaination.

UGC NET Entrance exams Science # Electronic Science Current Electricity Prepare / Learn

Q2. Find the pd between A and B from the figure

A. 4V

B. 2V

C. 6V

D. 8V

Answer : Option A
Explaination / Solution:

The resistance in the arm CAD RCAD = 4 + 6 = 10Omega .

The resistance in the arm CBD RCBD = 6 + 4 = 10Omega

Since the resistances in both the arms are equal, the current splits equally in the arms.ICAD = ICBD = 4/2 = 2A.

The potential difference across A and C.VAC = VA - VC = ICAD * RAC = 2 * 4 = 8V

potential difference across B and C VBC = VB - VC = ICBD * RBC = 2 * 6 = 12V.The potential difference across A and B VBA = VB - VA = VBC - VAC = 12 - 8 = 4V.

Point B is at a higher potential when compared to A.

UGC NET Entrance exams Science # Electronic Science Magnetism and Matter Prepare / Learn

Q3.

\vec{H}

=magnetic intensity,

χ

=susceptibility, magnetic moment per unit volume

\vec{M}

equals

A. 2χH⃗

B. χH⃗

C. χ2H⃗

D. πχH⃗

Answer : Option B
Explaination / Solution:
No Explaination.

UGC NET Entrance exams Science # Electronic Science Electronic Devices Prepare / Learn

Q4. The electron and hole concentration in a semiconductor in thermal equilibrium is given by

A. nenh=2n2i

B. nenh= n2i

C. 2nenh=n2i

D. nenh=n3i

Answer : Option B
Explaination / Solution:

The number of carriers in the conduction and valence band with no externally applied bias is called the equilibrium carrier concentration.

For majority carriers, the equilibrium carrier concentration is equal to the intrinsic carrier concentration plus the number of free carriers added by doping the semiconductor. Under most conditions, the doping of the semiconductor is several orders of magnitude greater than the intrinsic carrier concentration, such that the number of majority carriers is approximately equal to the doping.

At equilibrium, the product of the majority and minority carrier concentration is a constant, and this is mathematically expressed by the Law of Mass Action.

$n_{e} n_{h} = n_{i}^{2}$

Where ni is the intrinsic carrier concentration and neand ph are the electron and hole equilibrium carrier concentrations.

UGC NET Entrance exams Science # Electronic Science Power Electronics Prepare / Learn

Q5. The Voltage Source Inverter (VSI) shown in the figure below is switched to provide a 50 Hz, square-wave ac output voltage (v0) across an R-L load. Reference polarity of v0 and reference direction of the output current i0 are indicated in the figure. It is given that R = 3 Ω, L = 9.55 mH.

Appropriate transition i.e., Zero Voltage Switching (ZVS) / Zero Current Switching (ZCS) of the IGBTs during turn-on / turn-off is

A. ZVS during turn off

B. ZVS during turn-on

C. ZCS during turn off

D. ZCS during turn-on

Answer : Option D
Explaination / Solution:
No Explaination.

UGC NET Entrance exams Science # Electronic Science Digital Logic Prepare / Learn

Q6. Which one of the following expressions does NOT represent exclusive NOR of x and y?

A. xy + x 'y'

B. x ⊕ y'

C. x '⊕ y

D. x '⊕ y'

Answer : Option D
Explaination / Solution:

UGC NET Entrance exams Science # Electronic Science Analog Circuits Prepare / Learn

Q7. The following circuit has R = 10kΩ, C = 10μF . The input voltage is a sinusoid at 50Hz with an rms value of 10V. Under ideal conditions, the current is from the source is

A. 10 π mA leading by 90°

B. 10 π mA leading by 90°

C. 20 π mA leading by 90°

D. 10 π mA lagging by 90°

Answer : Option D
Explaination / Solution:
No Explaination.

UGC NET Entrance exams Science # Electronic Science Digital Circuits Prepare / Learn

Q8. A JK flip flop can be implemented by T flip-flops. Identify the correct implementation.

Answer : Option B
Explaination / Solution:

UGC NET Entrance exams Science # Electronic Science Electrostatic Potential and Capacitance Prepare / Learn

Q9. An equipotential surface is a surface

A. on which the potential has the zero value at every point

B. on which the potential has positive value at every point

C. on which the potential has negative value at every point

D. on which the potential has the same value at every point

Answer : Option D
Explaination / Solution:

When the potential at all points on the surface is the same; such a surface is called an equipotential surface. The potential difference between any two points on an equipotential surface is zero, while the potential may be positive, negative or even zero.

UGC NET Entrance exams Science # Electronic Science Moving Charges and Magnetism Prepare / Learn

Q10. The magnetic field at the center of a toroidal coil having n turns,radius r and carrying a current I in air is

A. μ0NI22πr

B. μ0NI
2πr

C. μ0N2I2πr

D. μ0NIπr

Answer : Option B
Explaination / Solution:

A toroid is a solenoid wound on itself. The magnetic field exists only in the region of the hollow circular ring which has the turns of wire wound on it. Using Ampere’s circuital law,

B (2 π r) = μ_{0} N I; B = \frac{μ_{0} N I}{2 π r}

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UGC NET Entrance exams Science Electronic Science TEST : 17