Question is ⇒ The circuit shown in figure, Options are ⇒ (A) oscillate at 456 kHz, (B) oscillate at 228 kHz, (C) oscillate at microwave frequency, (D) not oscillate, (E) , Leave your comments or Download question paper.

Question is ⇒ The circuit shown in figure, Options are ⇒ (A) oscillate at 456 kHz, (B) oscillate at 228 kHz, (C) oscillate at microwave frequency, (D) not oscillate, (E) , Leave your comments or Download question paper. Previous question Next question


Q1.	The circuit shown in figure
A.	oscillate at 456 kHz [Wrong Answer]
B.	oscillate at 228 kHz [Wrong Answer]
C.	oscillate at microwave frequency [Wrong Answer]
D.	not oscillate [Correct Answer]
View Answer 1 -3 Explanation:- Answer : D Discuss it below : !! OOPS Login [Click here] is required to post your answer/result Help other students, write article, leave your comments

Previous question Next question

Also Read Similar Questions Below :

⇒ Ferrite have
low copper loss
low eddy current loss
low resistivity
higher specific gravity compared to iron

⇒ An R-C coupled amplifier has mid-frequency gain of 200 and a frequency response from 100 Hz to 20 kHz. A negative feedback network with β = 0.2 is incorporated into the amplifier circuit, the Bandwidth will be
50 kHz
20 kHz
100 kHz
infinite

⇒ Two wattmeter method is suitable only for balanced loads.
TRUE
FALSE

⇒ A signal with ± 10 V ranged and 1 KHz band width is being digitized using a sample and hold circuit and a 10 bit quantizer. The quantization step is :
1 x 10^-9 V
5 x 10^-9 V
2.5 x 10^-9 V
0.5 x 10^-9 V

⇒

Assertion (A): Two transistors one n-p-n and the other p-n-p are identical in all respects (doping, construction, shape, size). The n-p-n transistor will have better frequency response.

Reason (R): The electron mobility is higher than hole mobility.

Both A and R are true and R is correct explanation of A
Both A and R are true but R is not a correct explanation of A
A is true but R is false
A is false but R is true

⇒ Compared to bipolar junction transistor, a JFET has
lower input impedance
high input impedance and high voltage gain
higher voltage gain
high input impedance and low voltage gain

⇒ The time base of a high frequency CRO whose screen diameter is 10 cm is set at 10 μ sec/cm. The lowest frequency which can be fully displayed on the screen is
more than 100 kHz
equal to 100 kHz
equal to 10 kHz
less than 10 kHz

⇒ A 7477 decoder drives
seven-segment display
nickel tape
relays
diode matrix

⇒ In an ordinary thyristor
turn on and turn off times are equal
turn on time is more than turn off time
turn off time is more than turn on time
turn on time may be equal, less than or more than turn off time

⇒ The energy of constant amplitude complex valued exponential sequence is ...
A²
∞
1
0

⇒ When two 4 bit parallel adders are cascaded we get
4 bit parallel adder
8 bit parallel adder
16 bit parallel adder
none of the above

⇒ In 8085, TRAP is
always maskable
cannot interrupt a service subroutine
used for catastrophic events like temporary power failure
lowest priority interrupt

⇒ The region between a pin of parallel perfectly conducting planes of infinite extent is y and z directions is partially filled with a dielectric as shown be
-2
2
0
1

⇒ For operation of enhancement only N-channel MOSFET, value of gate voltage has to be
zero
low positive
high negative
high positive

⇒ In a Hurwitz polynomial
all coefficients are non-negative
the coefficients may be positive or negative
all coefficients are negative
all coefficients are non-negative and all zeros should be either in left half plane or simple on the imaginary axis

⇒ 8255 A has three 8 bit ports A, B, C for peripheral data transfer. For programming 8255 A, these ports are grouped as
Port A with Port B
Port A with 4 most significant ports of C, port B with 4 least significant ports of C
Port A with Port C
none of the above

⇒ The decimal equivalent of the binary number 10110.0101011101 is
22.24082165
22.340821675
22.3408213125
22.3408203125

⇒ The conductivity of germanium increases by about 6 percent per degree increase in temperature.
TRUE
FALSE

⇒ In a lag network the input frequency is 100 f_c where f_c is critical frequency. The voltage gain is
0 dB
-20 dB
-40 dB
-60 dB

⇒ A circuit using an op-amp is shown in the given figure. It
voltage series feedback
voltage shunt feedback
current shunt feedback
current series feedback

⇒ What kind of signal does the times produce?
Single pulse
Square wave
Continuous pulse
No signal

⇒ For a binary half subtractor having two inputs A and B the correct sets of logical expressions for the outputs D (A - B) and X = Borrow are
D = AB + AB, X = AB
D = AB AB, X = AB
D = AB + AB, X = AB
D = AB + A B, X = AB

⇒ In a series ac circuit a voltage of 20 V at 25 Hz causes a current of 0.20 A, while the same voltage at 75 Hz causes a current of 0.12 A. The circuit consists of
R and C in parallel
R and C in series
R and L in series
either (a) or (b)

⇒ In waveguide Degenerate modes are
two mode having same cut off frequency
two mode having same power
two mode having same phase velocity
it is a hypothetical mode

⇒ FM broadcast lies in which band?
HF
SHF
UHF
VHF

⇒

Match the following:

	List I		List II
A.	Darlington amplifier	1.	A circuit using pnp and npn transistors
B.	CB amplifier	2.	A CS amplifier driving a CG amplifier
C.	Cascade amplifier	3.	A circuit with over-all voltage gain close to 1 and very large input impedance
D.	Complementary symmetry amplifier	4.	A circuit with low input impedance

A-1, B-2, C-3, D-4
A-2, B-1, C-3, D-4
A-3, B-4, C-2, D-1
A-4, B-3, C-2, D-1

⇒ Which one of the following cannot be used to remove the unwanted sideband in SSB?
Balanced modulator
Phase shift method
Filter system
Third method

⇒ If the intrinsic resistivity at 300°K is 0.37 ohm-m and the electron and hole mobilities at 300°K are 0.28 and 0.15 m²/volt-sec, then the intrinsic carrier concentration of Germanium is __________ .
2.92 x 10¹⁹/m³
2.92 x 10^-19/m³
1.92 x 10¹⁹/m³
3.92 x 10¹⁹/m³

⇒ The main drawbacks of EEPROM are
low density
high cost
both low density and high cost
none of the above

⇒ Which one condition is true to check the periodically for discrete time signal (where K is any integer, N is period, f₀ is frequency of signal)
f₀ = K/N
f₀ = K + N
f₀ = KN
f₀ = N/K