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In the circuit of figure both diodes are ideal. If v_{1} = 10 V and v_{2} = 10 V which diode will cond | |
A. | D_{1} only [Wrong Answer] |
B. | D_{2} only [Wrong Answer] |
C. | Both D_{1} and D_{2} [Correct Answer] |
D. | Neither D_{1} nor D_{2} [Wrong Answer] |
View Answer
Explanation:-
Answer : C Discuss it below : !! OOPS Login [Click here] is required to post your answer/result |
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⇒ In the circuit of figure the current through 5 Ω resistance at t = 0^{+0.1666666666666672.5 A3.1 A0}
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A-4, B-1, C-3, D-2
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A-2, B-3, C-1, D-4
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4 bit subtractor Q - P
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Both A and R are true but R is not correct explanation of A
A is true but R is false
A is false but R is true
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Match the following:
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C. | AF amplifier | 3. | Amplifier frequencies in the range of 0 Hz to 5 MHz |
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⇒ Two resistance R_{1} and R_{2} are connected in series R_{1} = 528 ± 5 Ω and R_{2} = 325 ± 3Ω. The total resistance will be
853 ± 2 Ω
853 ± 5 Ω
853 ± 3 Ω
853 ± 8 Ω
⇒ The expression Y = pM (0, 1, 3, 4) is
POS
SOP
Hybrid
none of the above
⇒ If sound intensity is 8 W/m^{2}, the dB level of this sound is about
5 dB
50 dB
129 dB
1290 dB
⇒ Flag bits in arithmetic unit provide
status type information
repeatability
facilities for rechecks
all of the above
⇒ Which of the following noise does not occur in transistors?
Partition noise
Shot noise
Flicker noise
Resistance noise
⇒ The inverse response of a system h(n) = a^{n}∪(n) what is the condition for the system to be BIBO stable?
a is real and +ve
a is real and -ve
|a| > 1
|a| < 1
⇒ The unit of thermal resistance of a semi-conductor device is
Ohms
Ohms/°C
°C/ohm
°C/watt
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Both A and R are correct and R is correct explanation for A
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has two stable state
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D. | > > | 4. | Not equal to |
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⇒ Magnetic deflection is inversely proportional to
voltage
(voltage)^{0.5}
(voltage)^{1.5}
(voltage)^{2}
⇒ The main application of enhancement mode MOSFET is in
amplifier circuits
oscillator circuits
switching circuits
all of the above
⇒ Three point tracking is achieved with
double conversion
padder capacitor
double spotting
variable selectivity
⇒ A variable resistance R and capacitive reactance X_{C} are fed by an voltage. The current locus is
a semi circle in 4th quadrant
a semi circle in first quadrant
a straight line in 4th quadrant
a straight line first quadrant
⇒ For matching over a range of frequencies in a transmission line it is best to use
a balun
a broad band directional coupler
double stub
a single stub of adjustable position
⇒
Assertion (A): A decrease in temperature increases the reverse saturation current in a p-n diode.
Reason (R): When a diode is reverse biased surface leakage current flows.
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
⇒ A pre-emphasis circuit provides extra noise immunity by
converting the phase modulation of FM
preamplifying the whole audio band
amplifying the higher audio frequencies
boosting the pass frequencies
⇒ The silicon sample with unit cross-sectional area shown below is in thermal equilibrium. The following information is given: T = 300 K, electronic charge = 1.6 x 10^{-19}C, thermal voltage = 26mV and electron mobility = 1350 cm^{2}
1 kV/cm
5 kV/cm
10 kV/cm
26 kV/cm
⇒ The symbol in figure is
UJT
PUT
SCS
SBS
⇒ Given I_{s} = 20 A, V_{s} = 20 V, the current I in the 3 Ω resistance is give
0.166666666666667
0.333333333333333
0.0833333333333333
16 A
⇒ The number of accumulators in 6800 are
2
3
1
4
⇒ In the circuit of fi
V_{a} would always lag V_{b}
V_{a} would always lead V_{b}
V_{a} and V_{b} would always be in phase
V_{a} may lag or lead V_{b}
⇒ A single phase full bridge inverter for R-L loads needs
4 thyristors
4 thyristors and 4 diodes
4 thyristors and 2 diodes
8 thyristors
⇒ The general purpose register code for accumulator in 8085 is
111
0
1
1111
⇒ A small signal source v_{i}(t) = A cos 20t + B sin10^{6}t is applied to a transistor amplifier as shown below. The transistor has β = 150 and h_{ie} = 3 kΩ. Which expression best approximates V_{0}(t<
v_{0}(t) = -1500 (A cos 20t + B sin 10^{6}t)
v_{0}(t) = -150 (A cos 20t + B sin 10^{6}t)
v_{0}(t) = -1500 B sin 10^{6}t
V_{0}(t) = -1500 sin 10^{6}t
⇒ If V_{a1}, V_{a2}, V_{a0} are the symmetrical components of V_{a}, then
V_{a} = V_{a1} + V_{a2} + V_{a0}
V_{a} = a^{2}V_{a1} + aV_{a2} + V_{a0}
V_{a} = aV_{a1} + a^{2}V_{a2} + V_{a0}
V_{a} = aV_{a0} + a^{2}V_{a1} + V_{a2}