Electronic Science - Online Test
Q1. An N-type Ge is obtained on doping the Ge- crystal with
Answer : Option C
Explaination / Solution:
The addition of pentavalent impurities such as antimony, arsenic or phosphorous contributes free electrons, greatly increasing the conductivity of the intrinsic semiconductor.
Q2.
The ratio of the mobility to the diffusion coefficient in a semiconductor has the units
Answer : Option A
Explaination / Solution:
Q3. The increasing order of speed of data access for the following devices is
i. Cache Memory
ii. CDROM
iii. Dynamic RAM
iv. Processor Registers
v. Magnetic Tape
Answer : Option B
Explaination / Solution:
No Explaination.
Q4. The electron and hole concentrations in an intrinsic semiconductor are ni per cm3 at 300 K. Now, if acceptor impurities are introduced with a concentration
of NA per cm3
(where NA >> ni , the electron concentration per cm3 at 300 K
will be
Answer : Option D
Explaination / Solution:
As per mass action law
np = ni
If acceptor impurities are introduces
Q5. In the given network V1 = 100∠0° V, V2 = 100∠ − 120° V, V3 = 100∠ + 120° V. The phasor
current 𝑖 (in Ampere) is
Answer : Option D
Explaination / Solution:
Q6. In the given network V1 = 100∠0°V, V2 = 100∠-120°V, V3 = 100∠+120°V. The phasor
current i (in Ampere) is
Answer : Option A
Explaination / Solution:
Q7. For a charged conductor of arbitrary shape, inside the conductor
Answer : Option C
Explaination / Solution:
The electric field on the surface of a hollow conductor is maximum and it drops to zero abruptly inside the conductor. Since, the potential difference between any two points inside the hollow conductor is zero. This means that the potential at all points inside the hollow charged conductor is same and it is equal to the value of the potential at its surface.
Q8. In the circuit shown, the op-amp has finite input impedance, infinite voltage
gain and zero input offset voltage. The output voltage Vout is
Answer : Option C
Explaination / Solution:
Given that the op-amp has infinite voltage gain, i.e.
AOL =
and zero input offset voltage
VIO = 0
So, we redraw the op-amp circuit as
Hence, the current I1 is drawn through resistance R2. So, the output voltage is
Vout = I1R2
Q9. Current density of a conductor is
Answer : Option B
Explaination / Solution:
Current density J = I/A
In electromagnetism, current density is the electric current per unit area of cross section. It is a vector and has a direction along the area vector.
Q10. A circular coil of radius R carries an electric current. The magnetic field due to the coil at a point on the axis of the coil located at a distance r from the center of the coil such that r>>R, varies as
Answer : Option B
Explaination / Solution:
At a point distance r from the coil, the magnetic field is . If is neglected in the denominator;
