Govt Exams Mock Test FREE | Online Quiz Competition

Q1. In voltage divider bias, \(V_{CC} = 25\: V;\: R_1 = 10\: k\Omega;\: R_2 = 3.3 \: k\Omega; \:R_C = 2 \: k\Omega;\) and \(R_E =1\: k\Omega\). What is the collector current?

6.2 mA

5.5 mA

4.5 mA

3.6 mA

Q2. In voltage divider bias, \(V_{CC} = 25\: V;\: R_1 = 10\: k\Omega;\: R_2 = 3.3 \: k\Omega; \:R_C = 2 \: k\Omega;\) and \(R_E =1\: k\Omega\). What is the emitter current?

6.2 mA

5.5 mA

4.5 mA

3.6 mA

Q3. An amplifier without feedback has a voltage gain of 100, input resistance of \(10\:K\Omega\) & Output resistance of \(3\:K\Omega\). The input resistance and output resistance of the voltage-series negative feedback amplifier using the above amplifier with a feedback factor of 0.6 is

\(\frac{10}{61}\: K\Omega\) and \(183\: K\Omega\)

\(\frac{10}{61}\: K\Omega\) and \(\frac{3}{61}\: K\Omega\)

\(0.6\: K\Omega\) and \(\frac{3}{6}\: K\Omega\)

\(610\: K\Omega\) and \(\frac{3}{61}\: K\Omega\)

Q4. In voltage divider bias, operating point is 3 V, 2 mA. If \(V_{CC} = 9\: V;\: R_C = 2.2\: k\Omega \). What is the value of \(R_E \)?

\(727 \Omega\)

\(1 k \Omega\)

\(930 \Omega\)

\(646 \Omega\)

Q5. Which type of BJT configuration is providing the maximum power gain?

Common Base

Common Collector

Common Emitter

Both common emitter and base

Q6. In analog computation, we prefer integrator to differentiator because

integrator is more stable

integrator has less voltage drift and noise

integrator is cheaper

both (A) and (B) above

Q7. Current flow in a semiconductors depends on the phenomenon of

drift

diffusion

recombination

all of these

Q8. For emitter bias, the voltage at the emitter is 0.7 V less than the

Base voltage.

Emitter voltage

Collector voltage

Ground voltage

Q9. A low-pass filter with a cut-off frequency of 30 Hz is cascaded with a high-pass filter with a cut-off frequency of 20Hz. The resultant system of filters will function as

An all-pass filter

An all-stop filter

An band stop (band-reject) filter

A band-pass filter

Q10. The relation between slew rates, frequency is given by

\(f_{max} = \frac {SR}{6.2 \ast V_m}\)

\(f_{max} = \frac {V_m}{6.2 \ast SR}\)

\(f_{max} = \frac {V_m \ast 6.2}{SR}\)

\(f_{max} = \frac {SR}{V_m}\), SR = Slew Rate

Quiz

Year 2024

September
Week # 3 Quiz 3

Q1. In voltage divider bias, \(V_{CC} = 25\: V;\: R_1 = 10\: k\Omega;\: R_2 = 3.3 \: k\Omega; \:R_C = 2 \: k\Omega;\) and \(R_E =1\: k\Omega\). What is the collector current?

Q2. In voltage divider bias, \(V_{CC} = 25\: V;\: R_1 = 10\: k\Omega;\: R_2 = 3.3 \: k\Omega; \:R_C = 2 \: k\Omega;\) and \(R_E =1\: k\Omega\). What is the emitter current?

Q3. An amplifier without feedback has a voltage gain of 100, input resistance of \(10\:K\Omega\) & Output resistance of \(3\:K\Omega\). The input resistance and output resistance of the voltage-series negative feedback amplifier using the above amplifier with a feedback factor of 0.6 is

Q4. In voltage divider bias, operating point is 3 V, 2 mA. If \(V_{CC} = 9\: V;\: R_C = 2.2\: k\Omega \). What is the value of \(R_E \)?

Q5. Which type of BJT configuration is providing the maximum power gain?

Q6. In analog computation, we prefer integrator to differentiator because

Q7. Current flow in a semiconductors depends on the phenomenon of

Q8. For emitter bias, the voltage at the emitter is 0.7 V less than the

Q9. A low-pass filter with a cut-off frequency of 30 Hz is cascaded with a high-pass filter with a cut-off frequency of 20Hz. The resultant system of filters will function as

Q10. The relation between slew rates, frequency is given by

Quiz

Year 2024

September Week # 2 Expire on 07 Oct 2024

September Week # 3 Expire on 14 Oct 2024

September Week # 4 Expire on 21 Oct 2024

October Week # 1 Expire on 31 Oct 2024

September Week # 3 Quiz 3

Q1. In voltage divider bias, \(V_{CC} = 25\: V;\: R_1 = 10\: k\Omega;\: R_2 = 3.3 \: k\Omega; \:R_C = 2 \: k\Omega;\) and \(R_E =1\: k\Omega\). What is the collector current?

Q2. In voltage divider bias, \(V_{CC} = 25\: V;\: R_1 = 10\: k\Omega;\: R_2 = 3.3 \: k\Omega; \:R_C = 2 \: k\Omega;\) and \(R_E =1\: k\Omega\). What is the emitter current?

Q3. An amplifier without feedback has a voltage gain of 100, input resistance of \(10\:K\Omega\) & Output resistance of \(3\:K\Omega\). The input resistance and output resistance of the voltage-series negative feedback amplifier using the above amplifier with a feedback factor of 0.6 is

Q4. In voltage divider bias, operating point is 3 V, 2 mA. If \(V_{CC} = 9\: V;\: R_C = 2.2\: k\Omega \). What is the value of \(R_E \)?

Q5. Which type of BJT configuration is providing the maximum power gain?

Q6. In analog computation, we prefer integrator to differentiator because

Q7. Current flow in a semiconductors depends on the phenomenon of

Q8. For emitter bias, the voltage at the emitter is 0.7 V less than the

Q9. A low-pass filter with a cut-off frequency of 30 Hz is cascaded with a high-pass filter with a cut-off frequency of 20Hz. The resultant system of filters will function as

Q10. The relation between slew rates, frequency is given by

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September
Week # 3 Quiz 3