$\mathrm{If} \mathrm{\alpha } \mathrm{be} \mathrm{the} \mathrm{current} \mathrm{gain} \mathrm{of} \mathrm{a} \mathrm{transistor} \mathrm{in} \mathrm{common} \mathrm{base} \mathrm{mode} \mathrm{and} \mathrm{\beta } \mathrm{be} \mathrm{the} \mathrm{current} \mathrm{gain} \mathrm{in}$
$\mathrm{common} \mathrm{emitter} \mathrm{mode} \mathrm{then}-$

1.  $\mathrm{\alpha } < 1$

2.  $\mathrm{\beta } > 1$

3.  $\mathrm{\alpha } = \frac{\mathrm{\beta }}{1 + \mathrm{\beta }}$

4.  All of these

 

The conductivity of an n-type semiconductor whose density of conduction electrons is ${\mathrm{n}}_{\mathrm{e}}$, Density of holes is ${\mathrm{n}}_{\mathrm{h}}$, Mobility of conduction electrons is ${\mathrm{m}}_{\mathrm{e}}$and Mobility of holes is ${\mathrm{m}}_{\mathrm{h}},$ will be

1.  $\mathrm{e}\left[{\mathrm{n}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{h}} + {\mathrm{n}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{e}}\right]$

2.  $\mathrm{e}\left[{\mathrm{n}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{e}} + {\mathrm{n}}_{\mathrm{h}}{\mathrm{m}}_{\mathrm{e}}\right]$

3.  ${\mathrm{en}}_{\mathrm{e}}\left[{\mathrm{m}}_{\mathrm{e}} + {\mathrm{m}}_{\mathrm{h}}\right]$

4.  $\mathrm{e}\left[{\mathrm{n}}_{\mathrm{e}}{\mathrm{m}}_{\mathrm{e}} + {\mathrm{n}}_{\mathrm{h}}{\mathrm{m}}_{\mathrm{h}}\right]$

The current through an ideal p-n junction diode shown in the circuit will be:

1.  5 A
2.  0.2 A 
3.  0.6 A
4.  zero

$\mathrm{In} \mathrm{the} \mathrm{given} \mathrm{circuit} \mathrm{power}$
$\mathrm{developed} \mathrm{in} 1\mathrm{k\Omega } \mathrm{resistor} \mathrm{is}$

1.  36 mW

2.  12 mW

3.  144 mW

4.  64 mW

A semiconductor is known to have an electron concentration of \(8\times 10^{13}~\text{cm}^{-3}\) and a hole concentration of \(5\times 10^{2}~\text{cm}^{-3}\). The semiconductor is:

Which of the following is correct for \(\mathrm{n}\)-type semiconductors?

When a transistor is used as a switch it is in:

1.  Active state

2.  Cut off state

3.  Saturation state

4.  Both cut off state and saturation state are possible

A transistor is operated in common emitter configuration at ${\mathrm{V}}_{\mathrm{c}} = 10\mathrm{V}.$ When base current is changed from 10mA to 30mA, it produces a change in emitter current from 2A to 4A, the current amplification factor is

1.  100

2.  99

3.  10

4.  9