1
MCQ (Single Correct Answer)

### JEE Main 2017 (Online) 8th April Morning Slot

The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is :

A
10
B
10$-$6
C
106
D
100

## Explanation

Forward bias resistance

R1 = ${{\Delta V} \over {\Delta I}}$ = ${{0.8 - 0.7} \over {\left( {20 - 10} \right) \times {{10}^{ - 3}}}}$ = 10 $\Omega$

Reverse bias resistance,

R2 = ${{10} \over {1 \times {{10}^{ - 6}}}}$ = 107 $\Omega$

$\therefore\,\,\,$ Ratio of resistance

= ${{{R_1}} \over {{R_2}}}$ = ${{10} \over {{{10}^7}}}$ = 10$-$6

2
MCQ (Single Correct Answer)

### JEE Main 2017 (Online) 9th April Morning Slot

The current gain of a common emitter amplifier is 69. If the emitter current is 7.0 mA, collector current is :
A
9.6 mA
B
6.9 mA
C
0.69 mA
D
69 mA

## Explanation

Here, $\beta$ = 69, Ie = 7 mA, Ic = ?

$\alpha$ = ${\beta \over {1 + \beta }}$ = ${{69} \over {70}}$

Also, $\alpha$ = ${{{I_c}} \over {{I_e}}}$

$\Rightarrow$ ${{69} \over {70}} = {{{I_c}} \over 7}$

$\Rightarrow$ Ic = ${{69} \over {70}} \times 7$ = 6.9 mA
3
MCQ (Single Correct Answer)

### JEE Main 2018 (Offline)

The reading of the ammeter for a silicon diode in the given circuit is :
A
13.5 mA
B
0
C
15 mA
D
11.5 mA

## Explanation

Note :

From the circuit you can see the diode is in forward bias, and to start current flow through the circuit. Accross silicon diode the voltage drop should be 0.7 V and for Germenium diode the voltage drop is 0.3 V. Then diode will start working and current will flow through the circuit.

So, the net voltage available accros the registor is (Vnet) = 3 $-$ 0.7 = 2.3 V

$\therefore\,\,\,$ Current flow through circuit = ${{2.3} \over {200}}$ = 11.5 mA
4
MCQ (Single Correct Answer)

### JEE Main 2018 (Online) 15th April Morning Slot

In a common emitter configuration with suitable bias, it is given that ${R_L}$ is the load resistance and ${R_{BE}}$ is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by :

$\beta$ is curret gain, ${{\rm I}_B},{{\rm I}_C}$ and ${{\rm I}_E}$ are respectively base, collector and emitter currents.
A
$\beta {{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_C}} \over {\Delta {{\rm I}_B}}},{\beta ^2}{{{R_L}} \over {{R_{BE}}}}$
B
$\beta {{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_E}} \over {\Delta {{\rm I}_B}}},{\beta ^2}{{{R_L}} \over {{R_{BE}}}}$
C
${\beta ^2}{{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_C}} \over {\Delta {{\rm I}_E}}},{\beta ^2}{{{R_L}} \over {{R_{BE}}}}$
D
${\beta ^2}{{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_C}} \over {\Delta {{\rm I}_B}}},\beta {{{R_L}} \over {{R_{BE}}}}$

## Explanation

Current gain ($\beta$) = ${{\Delta \,{I_C}} \over {\Delta {I_B}}}$

Voltage gain = ${{{V_{CE}}} \over {{V_{BE}}}} = \beta {{{R_L}} \over {{R_{BE}}}}$

Power gain = voltage gain x current gain = ${\beta ^2}{{{R_L}} \over {{R_{BE}}}}$

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