Explore topic-wise MCQs in Linear Integrated Circuit.

This section includes 6 Mcqs, each offering curated multiple-choice questions to sharpen your Linear Integrated Circuit knowledge and support exam preparation. Choose a topic below to get started.

1.

For a full wave rectification, in a low voltage ac voltmeter, the meter current can be expressed a?

A. I<sub>o</sub> = (1.9√óV<sub>in</sub>)/R<sub>1</sub>
B. I<sub>o</sub> = (3.9√óV<sub>in</sub>)/R<sub>1</sub>
C. I<sub>o</sub> = (0.9√óV<sub>in</sub>)/R<sub>1</sub>
D. I<sub>o</sub> = (2.9√óV<sub>in</sub>)/R<sub>1</sub>
Answer» D. I<sub>o</sub> = (2.9‚Äö√†√∂‚àö‚â•V<sub>in</sub>)/R<sub>1</sub>
2.

The current to voltage converter photosensitive device can be used as

A. Light intensity meter
B. Light radiating meter
C. Light deposition meter
D. None of the mentioned
Answer» B. Light radiating meter
3.

If the input applied to DAC using current to voltage converter is 10110100, determine the reference voltage (Assume Io= 2mA and R1=1.2kΩ)$

A. 53.1v
B. 3.41v
C. 9.21v
D. 67.34v
Answer» C. 9.21v
4.

Which cell can be used instead of a photocell to obtain active transducer in photosensitive devices?

A. Photovoltaic cell
B. Photo diode
C. Photo sensor
D. All of the mentioned
Answer» B. Photo diode
5.

Determine the maximum value of output current of the DAC in MC1408?

A. 0.773√ó(V<sub>ref</sub>/R<sub>1</sub>)
B. 0.448√ó(V<sub>ref</sub>/R<sub>1</sub>)
C. 0.996√ó(V<sub>ref</sub>/R<sub>1</sub>)
D. 0.224√ó(V<sub>ref</sub>/R<sub>1</sub>)
Answer» D. 0.224‚Äö√†√∂‚àö‚â•(V<sub>ref</sub>/R<sub>1</sub>)
6.

The output current equation for MC1408 digital to analog converter would be

A. I<sub>o</sub>= -(V<sub>ref</sub>/R<sub>1</sub>)√ó[(D<sub>7</sub>/2)+(D<sub>6</sub>/4)+(D<sub>5</sub>/8)+(D<sub>4</sub>/16)+(D<sub>3</sub>/32)+(D<sub>2</sub>/64)+(D<sub>1</sub>/128)+(D<sub>0</sub>/256)].
B. I<sub>o</sub>= (V<sub>ref</sub>/2R<sub>1</sub>)√ó[(D<sub>7</sub>/2)+(D<sub>6</sub>/4)+(D<sub>5</sub>/8)+(D<sub>4</sub>/16)+(D<sub>3</sub>/32)+(D<sub>2</sub>/64)+(D<sub>1</sub>/128)+(D<sub>0</sub>/256)].
C. I<sub>o</sub>= (V<sub>ref</sub>/R<sub>1</sub>)√ó[(D<sub>7</sub>/2)+(D<sub>6</sub>/4)+(D<sub>5</sub>/8)+(D<sub>4</sub>/16)+(D<sub>3</sub>/32)+(D<sub>2</sub>/64)+(D<sub>1</sub>/128)+(D<sub>0</sub>/256)].
D. I<sub>o</sub>= -(V<sub>ref</sub>/2R<sub>1</sub>)√ó[(D<sub>7</sub>/2)+(D<sub>6</sub>/4)+(D<sub>5</sub>/8)+(D<sub>4</sub>/16)+(D<sub>3</sub>/32)+(D<sub>2</sub>/64)+(D<sub>1</sub>/128)+(D<sub>0</sub>/256)].
Answer» D. I<sub>o</sub>= -(V<sub>ref</sub>/2R<sub>1</sub>)‚Äö√†√∂‚àö‚â•[(D<sub>7</sub>/2)+(D<sub>6</sub>/4)+(D<sub>5</sub>/8)+(D<sub>4</sub>/16)+(D<sub>3</sub>/32)+(D<sub>2</sub>/64)+(D<sub>1</sub>/128)+(D<sub>0</sub>/256)].