MCQOPTIONS
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MCQOPTIONS
This section includes 15 Mcqs, each offering curated multiple-choice questions to sharpen your Network Theory knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
In the following circuit. Find the current through 4Ω resistor using Millman’s Theorem. |
| A. | 0.5 |
| B. | 1 |
| C. | 1.5 |
| D. | 2 |
| Answer» D. 2 | |
| 2. |
Consider the circuit shown below. Find the current through 4Ω resistor. |
| A. | 2 |
| B. | 1.5 |
| C. | 1 |
| D. | 0.5 |
| Answer» C. 1 | |
| 3. |
Find the current through 3Ω resistor in the circuit shown below using Millman’s Theorem. |
| A. | 4 |
| B. | 3 |
| C. | 2 |
| D. | 1 |
| Answer» C. 2 | |
| 4. |
Calculate the current through 3Ω resistor in the circuit shown below. |
| A. | 1 |
| B. | 2 |
| C. | 3 |
| D. | 4 |
| Answer» D. 4 | |
| 5. |
According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then the value of equivalent conductance is? |
| A. | G’=R1+R2+⋯Rn |
| B. | G’=1/(1/R1+1/R2+⋯1/Rn) |
| C. | G’=1/((R1+R2+⋯Rn)) |
| D. | G’=1/R1+1/R2+⋯1/Rn |
| Answer» D. G’=1/R1+1/R2+⋯1/Rn | |
| 6. |
According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then the value of equivalent current source is? |
| A. | I‘=((I1R1+I2R2+⋯.+InRn))/(R1+R2+⋯Rn) |
| B. | I’=I1R1+I2R2+⋯.+InRn |
| C. | I’=((I1/R1+I2/R2+⋯.+In/Rn))/(R1+R2+⋯Rn) |
| D. | I’=I1/R1+I2/R2+⋯.+In/Rn |
| Answer» B. I’=I1R1+I2R2+⋯.+InRn | |
| 7. |
According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then these sources are replaced by? |
| A. | single voltage source V’ in parallel with G’ |
| B. | single current source I’ in series with G’ |
| C. | single current source I’ in parallel with G’ |
| D. | single voltage source V’ in series with G’ |
| Answer» D. single voltage source V’ in series with G’ | |
| 8. |
According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then the value of equivalent resistance is? |
| A. | R’=G1+G2+⋯Gn |
| B. | R’=1/G1+1/G2+⋯1/Gn |
| C. | R’=1/((G1+G2+⋯Gn)) |
| D. | R’=1/(1/G1+1/G2+⋯1/Gn) |
| Answer» D. R’=1/(1/G1+1/G2+⋯1/Gn) | |
| 9. |
According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then the value of equivalent voltage source is? |
| A. | V‘=(V1G1+V2G2+⋯.+VnGn) |
| B. | V‘=((V1G1+V2G2+⋯.+VnGn))/((1/G1+1/G2+⋯1/Gn)) |
| C. | V‘=((V1G1+V2G2+⋯.+VnGn))/(G1+G2+⋯Gn) |
| D. | V‘=((V1/G1+V2/G2+⋯.+Vn/Gn))/( G1+G2+⋯Gn) |
| Answer» D. V‘=((V1/G1+V2/G2+⋯.+Vn/Gn))/( G1+G2+⋯Gn) | |
| 10. |
According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then these sources are replaced by? |
| A. | single current source I’ in series with R’ |
| B. | single voltage source V’ in series with R’ |
| C. | single current source I’ in parallel to R’ |
| D. | single voltage source V’ in parallel to R’ |
| Answer» C. single current source I’ in parallel to R’ | |
| 11. |
In_the_circuit_shown_in_the_question_9_find_the_current_through_4Ω_resistor_using_Millman’s_Theorem.$# |
| A. | 0.5 |
| B. | 1 |
| C. | 1.5 |
| D. | 2 |
| Answer» D. 2 | |
| 12. |
Find the current through 3Ω resistor in the circuit shown above using Millman’s Theorem?# |
| A. | 4 |
| B. | 3 |
| C. | 2 |
| D. | 1 |
| Answer» C. 2 | |
| 13. |
According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then these sources are replaced by?$ |
| A. | single voltage source V’ in parallel with G’ |
| B. | single current source I’ in series with G’ |
| C. | single current source I’ in parallel with G’ |
| D. | single voltage source V’ in series with G’ |
| Answer» D. single voltage source V‚Äö√Ñ√∂‚àö√ë‚àö¬• in series with G‚Äö√Ñ√∂‚àö√ë‚àö¬• | |
| 14. |
In the question above, the value of equivalent voltage source is? |
| A. | V<sup>‘</sup>=(V<sub>1</sub>G<sub>1</sub>+V<sub>2</sub>G<sub>2</sub>+⋯.+V<sub>n</sub>G<sub>n</sub>) |
| B. | V<sup>‘</sup>=((V<sub>1</sub>G<sub>1</sub>+V<sub>2</sub>G<sub>2</sub>+⋯.+V<sub>n</sub>G<sub>n</sub>))/((1/G<sub>1</sub>+1/G<sub>2</sub>+⋯1/G<sub>n</sub>)) |
| C. | V<sup>‘</sup>=((V<sub>1</sub>G<sub>1</sub>+V<sub>2</sub>G<sub>2</sub>+⋯.+V<sub>n</sub>G<sub>n</sub>))/(G<sub>1</sub>+G<sub>2</sub>+⋯G<sub>n</sub>) |
| D. | V<sup>‘</sup>=((V<sub>1</sub>/G<sub>1</sub>+V<sub>2</sub>/G<sub>2</sub>+⋯.+V<sub>n</sub>/G<sub>n</sub>))/( G<sub>1</sub>+G<sub>2</sub>+⋯G<sub>n</sub>) |
| Answer» D. V<sup>‚Äö√Ñ√∂‚àö√ë‚àö‚â§</sup>=((V<sub>1</sub>/G<sub>1</sub>+V<sub>2</sub>/G<sub>2</sub>+‚Äö√Ñ√∂‚àö¬£‚àö√≤.+V<sub>n</sub>/G<sub>n</sub>))/( G<sub>1</sub>+G<sub>2</sub>+‚Äö√Ñ√∂‚àö¬£‚àö√≤G<sub>n</sub>) | |
| 15. |
According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then these sources are replaced by? |
| A. | single current source I’ in series with R’ |
| B. | single voltage source V’ in series with R’ |
| C. | single current source I’ in parallel to R’ |
| D. | single voltage source V’ in parallel to R’ |
| Answer» C. single current source I‚Äö√Ñ√∂‚àö√ë‚àö¬• in parallel to R‚Äö√Ñ√∂‚àö√ë‚àö¬• | |