1

### AIPMT 2002

For a transistor ${{{I_C}} \over {{I_E}}}$ = 0.96, then current gain for common emitter is
A
12
B
6
C
48
D
24

## Explanation

${{{I_C}} \over {{I_E}}}$ = 0.96 = $\alpha$

$\therefore$ Current gain, $\beta$ = ${{{I_C}} \over {{I_B}}}$ = ${\alpha \over {1 - \alpha }}$ = 24
2

### 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
3

### 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

### 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