# For the common emitter circuit as shown below if RC is changed from 2 KΩ, to 1 KΩ, and the transistor (β = 200) is replaced by anathor transistor ( β = 200 ) is replaced by another transistor (β = 100 ) then mode of operation of this transistor compared to earlier transistor is

1.  Active to saturation

2.  cutoff to active

3.  saturation to active

4.  remains same

4

remains same

Explanation :
No Explanation available for this question

# The voltage gain Av = vc/ vbe of the amplifier is

1.  385 V/V

2.  -385 V/V

3.   409 V/V

4.   -409 V/V

4

-385 V/V

Explanation :
No Explanation available for this question

# For the non-inverting integrator circuit shown below, if R1C1 = R2 C2, then it’s transfer function

1.  H(s) = 1/SR1C1

2.  H(s) = 1/SR2C2

3.  Both  a & b

4.  None

4

Both  a & b

Explanation :
No Explanation available for this question

# For the circuit shown below, assume β = 200, the value of output l voltage V0 & power dissipated by the transistor when input signal voltage Vi = 0.2V

1.  0V, 0 watt

2.  0V, 5 watt

3.  5V, 0 watt

4.  None

4

5V, 0 watt

Explanation :
No Explanation available for this question

# If the unregulated voltage increases by 20% the power dissipation across the transistor Q1 is

1.  Increases by 20%

2.  Increases by 50%

3.  Remains unchanged

4.  Decreases by 20%

4

Increases by 50\%

Explanation :
No Explanation available for this question

# In the above problem, if the gain is 20 dB, at 100Hz then, which of the following represents correct value of R2 & C2 respectively

1.  1 KΩ,159.16µF

2.  1 KΩ, 10 µF

3.  10 KΩ,1Kµ

4.  10 KΩ, 159.16μF

4

1 KΩ, 10 µF

Explanation :
No Explanation available for this question

# The VGG & Rs of the amplifier is

1.  0V, 20kΩ

2.  18V, 20kΩ

3.  13V, 18kΩ

4.  13V, 3.3kΩ

4

18V, 20kΩ

Explanation :
No Explanation available for this question

# For the circuit shown below, let β = 50, the value of input signal voltage V1 & power dissipated in the transistor when VBC = 0V is

1.  V1 = 0.818 V, PD = 6.46 mw

2.  V1 = 0.52 V, PD = 18.46 mw

3.  V1 = 0.52 V, PD = 6.98 mw

4.  V1 = 0.825V, PD = 6.98 mw

4

V1 = 0.825V, PD = 6.98 mw

Explanation :
No Explanation available for this question

# The range of possible value in ID if Rs = 3.3K and VGG = 0.

1.  0mA & 1.6 mA

2.  0.4 mA & 1.6 mA

3.  0.4 mA & 1.2 mA

4.  0.8 mA & 1.6 mA

4

0.4 mA & 1.2 mA

Explanation :
No Explanation available for this question

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