481.For a p-type semiconductor, which of the following statements is true? [NEET 2019]- (1)Holes are the majority carriers and trivalent atoms are the dopants.
- (2)Holes are the majority carriers and pentavalent atoms are the dopants.
- (3)Electrons are the majority carriers and pentavalent atoms are the dopants.
- (4)Electrons are the majority carriers and trivalent atoms are the dopants.
482.In the circuit shown here, assuming threshold voltage of diode is negligibly small, then voltage VAB is correctly represented by:- (1)VAB would be zero at all times
- (2)
- (3)
- (4)
483.The truth table for the circuit given below is:- (1)A
B Y 0 0 0 1 0 0 1 1 0 0 1 1 - (2)A
B Y 0 0 0 1 1 1 1 0 1 0 1 1 - (3)A
B Y 0 0 0 1 0 1 0 1 0 1 1 0 - (4)A
B Y 0 0 0 0 1 1 1 0 1 1 1 0
484.In the logic circuit given below, if X=1 and Y=1 then the values of P,Q and R are- (1)P=1,Q=1,R=0
- (2)P=0,Q=1,R=0
- (3)P=1,Q=0,R=1
- (4)P=1,Q=1,R=1
485.A B E 0 0 0 0 1 X 1 0 Y 1 1 0 In the truth table of the above circuit the value of X and Y are :- (1)0,0
- (2)1,1
- (3)1,0
- (4)0,1
486.The values of Y and Z in the given logic circuit are- (1)Y=1,Z=1
- (2)Y=0,Z=1
- (3)Y=1,Z=0
- (4)Y=0,Z=0
487.The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at Vz=5V and the desired current in load is 5 mA . The unregulated voltage source can supply upto 25 V . Considering the Zener diode can withstand four times of the load current, the value of resistor RS (shown in circuit) should be ____ Ω.- (1)100
- (2)1000
- (3)10
- (4)None of the above
488.When A=0 and B=1, the output is 0 for- (1)AND gate
- (2)OR gate
- (3)X-OR gate
- (4)NAND gate
489.For the given logic gates combination, the correct truth table will be- (1)A
B X 0 0 1 0 1 0 1 0 0 1 1 0 - (2)A
B X 0 0 0 0 1 1 1 0 1 1 1 0 - (3)A
B X 0 0 1 0 1 0 1 0 1 1 1 0 - (4)A
B X 0 0 0 0 1 1 1 0 1 1 1 1
490.The output (Y) of logic circuit given below is 0 only when:- (1)A=1,B=0
- (2)A=0,B=1
- (3)A=0,B=0
- (4)A=1,B=1
491.The correct truth table for the following logic circuit is :- (1)A
B Y 0 0 1 0 1 1 1 0 0 1 1 1 - (2)A
B Y 0 0 0 0 1 1 1 0 0 1 1 1 - (3)A
B Y 0 0 1 0 1 1 1 0 0 1 1 0 - (4)A
B Y 0 0 0 0 1 0 1 0 0 1 1 1
492.In the combination of the following gates the output Y can be written in terms of inputs A and B as: [2018]- (1)A⋅B
- (2)A⋅B+A⋅B
- (3)A⋅B'+A'⋅B
- (4)A+B
493.A p-n junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit.
The current (I) in the resistor (R) can be shown by:- (1)
- (2)
- (3)
- (4)
494.In the given circuit, when A=1,B=1 the values of X and Y respectively are- (1)1,0
- (2)1,1
- (3)0,1
- (4)0,0
495.Identify the logic operation carried out.- (1)OR
- (2)AND
- (3)NOR
- (4)NAND
496.The output (X) of the logic circuit shown in figure will be[2013]- (1)X=A,B
- (2)X=A'⋅B
- (3)X=A⋅B
- (4)X=A+B
497.The output of the circuit is low (zero) for: (A) X=0,Y=0 (B) X=0,Y=1 (C) X=1,Y=0 (D) X=1,Y=1 Choose the correct answer from the options given below:- (1)(B), (C) and (D) only
- (2)(A), (B) and (C) only
- (3)(A), (C) and (D) only
- (4)(A), (B) and (D) only
498.The output Y for the inputs A and B of circuit is given by
Truth table of the shown circuit is :- (1)A
B Y 0 0 1 0 1 1 1 0 1 1 1 0 - (2)A
B Y 0 0 1 0 1 0 1 0 0 1 1 1 - (3)A
B Y 0 0 0 0 1 1 1 0 1 1 1 1 - (4)A
B Y 0 0 0 0 1 1 1 0 1 1 1 0
499.Output of following logic circuit is- (1)(A+B)+(A+C)+(B+C)
- (2)(A+B)⋅(A+C)⋅(B+C)
- (3)(A⋅B)+(A⋅C)+(B⋅C)
- (4)None
500.The symbol given below represents- (1)A p-n-p transistor
- (2)An n-p-n transistor
- (3)A p-n junction diode
- (4)An inductor
501.As per the given circuit, the value of current through the battery will be...........A.
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