1
MCQ (Single Correct Answer)

AIPMT 2001

The current in the circuit will be

A
5/40 A
B
5/50 A
C
5/10 A
D
5/20 A

Explanation

D1 is reverse biased and D2 is forward biased.

$$ \therefore $$ I = $${5 \over {30 + 20}}$$ = $${5 \over {50}}$$ A
2
MCQ (Single Correct Answer)

AIPMT 2001

For a common base circuit if $${{{I_C}} \over {{I_E}}}$$ = 0.98 then current gain for common emitter circuit will be
A
49
B
98
C
4.9
D
25.5.

Explanation

$${{{I_C}} \over {{I_E}}}$$ = 0.98 = $$\alpha $$

$$ \therefore $$ Current gain, $$\beta $$ = $${{{I_C}} \over {{I_B}}}$$ = $${\alpha \over {1 - \alpha }}$$ = 49
3
MCQ (Single Correct Answer)

AIPMT 2001

The given truth table is for which logic gate

              $$\matrix{ A & B & Y \cr 1 & 1 & 0 \cr 0 & 1 & 1 \cr 1 & 0 & 1 \cr 0 & 0 & 1 \cr } $$
A
NAND
B
XOR
C
NOR
D
OR

Explanation

This truth table represents NAND gate.
4
MCQ (Single Correct Answer)

AIPMT 2000

The correct relation for $$\alpha $$, $$\beta $$ for a transistor
A
$$\beta = {{1 - \alpha } \over \alpha }$$
B
$$\beta = {\alpha \over {1 - \alpha }}$$
C
$$\alpha = {{\beta - 1} \over \beta }$$
D
$$\alpha \beta = 1$$

Explanation

$$\beta $$ = $${{{I_C}} \over {{I_B}}}$$ = $${{{I_C}} \over {{I_E} - {I_C}}}$$ = $${{{{{I_C}} \over {{I_E}}}} \over {1 - {{{I_C}} \over {{I_E}}}}}$$ = $${\alpha \over {1 - \alpha }}$$

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