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This section includes 16 Mcqs, each offering curated multiple-choice questions to sharpen your Electronic Devices Circuits knowledge and support exam preparation. Choose a topic below to get started.
1. |
Consider the figure given below. If the resistance R1 is disconnected from the ground and connected to a third power source v3, then expression for the value ofv0 is |
A. | 2v1 + 4v2 − 3v3 |
B. | 6v1 + 8v2 − 3v3 |
C. | 6v1 + 4v2 − 9v3 |
D. | 3v1 + 4v2 − 3v3 |
Answer» D. 3v1 + 4v2 − 3v3 | |
2. |
It is required to connect a transducer having an open-circuit voltage of 1 V and a source resistance of 1 MΩ to a load of 1-kΩ resistance. Find the load voltage if the connection is done (a) directly and (b) through a unity-gain voltage follower. |
A. | directly and (b) through a unity-gain voltage follower.a) 1 μV and 1 mV respectively |
B. | through a unity-gain voltage follower.a) 1 μV and 1 mV respectivelyb) 1 mV and 1 V respectively |
C. | 0.1 μV and 0.1 mV respectively |
D. | 0.1 mV and 0.1 V respectively |
Answer» C. 0.1 μV and 0.1 mV respectively | |
3. |
For designing a non-inverting amplifier with a gain of 2 at the maximum output voltage of 10 V and the current in the voltage divider is to be 10 μA the resistance required are R1 and R2 where R2 is used to provide negative feedback. Then |
A. | R1 = 0.5 MΩ and R2 = 0.5 MΩ |
B. | R1 = 0.5 kΩ and R2 = 0.5 kΩ |
C. | R1 = 5 MΩ and R2 = 5 MΩ |
D. | R1 = 5 kΩ and R2 = 5 kΩ |
Answer» B. R1 = 0.5 kΩ and R2 = 0.5 kΩ | |
4. |
The finite voltage gain of a non-inverting operational amplifier is A and the resistance used is R1 and R2 in which R2 is the feedback resistance. Under what conditions it can one use the expression 1 + R2/R1 to determine the gain of the amplifier? |
A. | A ~ R2/R1 |
B. | A >> R2/R1 |
C. | A << R2/R1 |
D. | None of the mentioned |
Answer» C. A << R2/R1 | |
5. |
While performing an experiment to determine the gain for an ideal operational amplifier having finite gain, a student mistakenly used the equation 1 + R2/R1 where R2 is the feedback resistance. What is the percentage error in his result? Given A is the finite voltage gain of the ideal amplifier used. |
A. | (R2/R1)/(A+ R2/R1) X 100% |
B. | (1+R2/R1)/(A+R2/R1) X 100% |
C. | (1+R2/R1)/(A+1+R2/R1) X 100% |
D. | (R2/R1)/(A+1+R2/R1) X 100% |
Answer» D. (R2/R1)/(A+1+R2/R1) X 100% | |
6. |
The gain for an ideal non-inverting operational amplifier is (given R2 is the feedback resistance) |
A. | R2/R1 – 1 |
B. | R2/R1 |
C. | -R2/R1 |
D. | R2/R1 + 1 |
E. | R2/R1 – 1b) R2/R1c) -R2/R1d) R2/R1 + 1 |
Answer» E. R2/R1 – 1b) R2/R1c) -R2/R1d) R2/R1 + 1 | |
7. |
For an ideal non-inverting operational amplifier having finite gain (A), the ratio of output voltage (v0) to input voltage (vi) is (given R2 is the feedback resistance)a) (1+R2/R1)/(1+((1+R2/R1)/A))b) (R2/R1)/(((1+R2/R1)/A))c) (1+R2/R1)/(((1+R2/R1)/A))d) (R2/R1)/(1+((1+R2/R1)/ |
A. | (1+R2/R1)/(1+((1+R2/R1)/A)) |
B. | (R2/R1)/(((1+R2/R1)/A)) |
C. | (1+R2/R1)/(((1+R2/R1)/A)) |
D. | (R2/R1)/(1+((1+R2/R1)/A)) |
E. | (1+R2/R1)/(1+((1+R2/R1)/A))b) (R2/R1)/(((1+R2/R1)/A))c) (1+R2/R1)/(((1+R2/R1)/A))d) (R2/R1)/(1+((1+R2/R1)/A)) |
Answer» B. (R2/R1)/(((1+R2/R1)/A)) | |
8. |
IT_IS_REQUIRED_TO_CONNECT_A_TRANSDUCER_HAVING_AN_OPEN-CIRCUIT_VOLTAGE_OF_1_V_AND_A_SOURCE_RESISTANCE_OF_1_M‚ÂÀ√≠¬¨¬©_TO_A_LOAD_OF_1-K‚ÂÀ√≠¬¨¬©_RESISTANCE._FIND_THE_LOAD_VOLTAGE_IF_THE_CONNECTION_IS_DONE_(A)_DIRECTLY_AND_(B)_THROUGH_A_UNITY-GAIN_VOLTAGE_FOLLOWER.?$# |
A. | 1 μV and 1 mV respectively |
B. | 1 mV and 1 V respectively |
C. | 0.1 μV and 0.1 mV respectively |
D. | 0.1 mV and 0.1 V respectively |
Answer» C. 0.1 ‚âà√≠¬¨‚à´V and 0.1 mV respectively | |
9. |
For designing a non-inverting amplifier with a gain of 2 at the maximum output voltage of 10 V and the current in the voltage divider is to be 10 μA the resistance required are R1 and R2 where R2 is used to provide negative feedback. The?# |
A. | R<sub>1</sub> = 0.5 MΩ and R<sub>2</sub> = 0.5 MΩ |
B. | R<sub>1</sub> = 0.5 kΩ and R<sub>2</sub> = 0.5 kΩ |
C. | R<sub>1</sub> = 5 MΩ and R<sub>2</sub> = 5 MΩ |
D. | R<sub>1</sub> = 5 kΩ and R<sub>2</sub> = 5 kΩ |
Answer» B. R<sub>1</sub> = 0.5 k‚âà√≠¬¨¬© and R<sub>2</sub> = 0.5 k‚âà√≠¬¨¬© | |
10. |
Which of the following is not true for a voltage follower amplifier? |
A. | Input voltage is equal to output voltage |
B. | Input resistance is infinite and output resistance is zero |
C. | It has 100% negative feedback |
D. | None of the mentioned |
Answer» E. | |
11. |
The finite voltage gain of a non-inverting operational amplifier is A and the resistance used is R1 and R2 in which R2 is the feedback resistance. Under what conditions it can one use the expression 1 + R2/R1 to determine the gain of the amplifier? |
A. | A ~ R<sub>2</sub>/R<sub>1</sub> |
B. | A >> R<sub>2</sub>/R<sub>1</sub> |
C. | A << R<sub>2</sub>/R<sub>1</sub> |
D. | None of the mentioned |
Answer» C. A << R<sub>2</sub>/R<sub>1</sub> | |
12. |
While performing an experiment to determine the gain for an ideal operational amplifier having finite gain, a student mistakenly used the equation 1 + R2/R1 where R2 is the feedback resistance. What is the percentage error in his result? Given A is the finite voltage gain of the ideal amplifier used. |
A. | (R2/R1)/(A+ R2/R1) X 100% |
B. | (1+R2/R1)/(A+R2/R1) X 100% |
C. | (1+R2/R1)/(A+1+R2/R1) X 100% |
D. | (R2/R1)/(A+1+R2/R1) X 100% |
Answer» D. (R2/R1)/(A+1+R2/R1) X 100% | |
13. |
The gain for an ideal non-inverting operational amplifier is (given R2 is the feedback resistance) |
A. | R<sub>2</sub>/R<sub>1</sub> – 1 |
B. | R<sub>2</sub>/R<sub>1</sub> |
C. | -R<sub>2</sub>/R<sub>1</sub> |
D. | R<sub>2</sub>/R<sub>1</sub> + 1 |
Answer» E. | |
14. |
For an ideal non-inverting operational amplifier having finite gain (A), the ratio of output voltage (v0) to input voltage (vi) is (given R2 is the feedback resistance) |
A. | (1+R2/R1)/(1+((1+R2/R1)/A)) |
B. | (R2/R1)/(((1+R2/R1)/A)) |
C. | (1+R2/R1)/(((1+R2/R1)/A)) |
D. | (R2/R1)/(1+((1+R2/R1)/A)) |
Answer» B. (R2/R1)/(((1+R2/R1)/A)) | |
15. |
For ideal non-inverting operational amplifier |
A. | Input and output resistances are infinite |
B. | Input resistance is infinite and output resistance is zero |
C. | Input resistance is zero and output resistance is infinite |
D. | Input and output resistances are zero |
Answer» C. Input resistance is zero and output resistance is infinite | |
16. |
In the non-inverting configuration of operational amplifier |
A. | The positive terminal is connected to the ground directly |
B. | The negative terminal is connected to the ground directly |
C. | The positive terminal is connected to the power source |
D. | The negative terminal is connected to the power source |
Answer» D. The negative terminal is connected to the power source | |