MCQOPTIONS
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MCQOPTIONS
This section includes 18 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. |
Let us assume x (t) = A cos(ωt + φ), What is the final steady state solution for y (t)? |
| A. | A|H(jω)|cos[ωt+Ø+ θ (ω)] |
| B. | A|H(jω)|cos[ωt-Ø+ θ (ω)] |
| C. | A|H(jω)|cos[ωt-Ø- θ (ω)] |
| D. | A|H(jω)|cos[ωt+Ø- θ (ω)] |
| Answer» B. A|H(jω)|cos[ωt-Ø+ θ (ω)] | |
| 2. |
Let us assume x (t) = A cos(ωt + φ), what is the value of k1 by considering θ (ω) is? |
| A. | |H(jω)|ej[θ (ω)+Ø] |
| B. | A/2|H(jω)|ej[θ (ω)+Ø] |
| C. | |H(jω)|e-j[θ (ω)+Ø] |
| D. | A/2 |H(jω)|e-j[θ (ω)+Ø] |
| Answer» C. |H(jω)|e-j[θ (ω)+Ø] | |
| 3. |
The relation between H (jω) and θ (ω) is? |
| A. | H(jω)=e-jθ (ω) |
| B. | H(jω)=|H(jω)|e-jθ (ω) |
| C. | H(jω)=|H(jω)|ejθ (ω) |
| D. | H(jω)=ejθ (ω) |
| Answer» D. H(jω)=ejθ (ω) | |
| 4. |
Let us assume x (t) = A cos(ωt + φ), what is the value of k1? |
| A. | 1/2 H(jω)AejØ |
| B. | H(jω)Ae-jØ |
| C. | H(jω)AejØ |
| D. | 1/2 H(jω)Ae-jØ |
| Answer» E. | |
| 5. |
Let us assume x (t) = A cos(ωt + φ), on taking the partial fractions for the response we get? |
| A. | Y(s)=k1/(s-jω)+(k1‘)/(s+jω)+Σterms generated by the poles of H(s) |
| B. | Y(s)=k1/(s+jω)+(k1‘)/(s+jω)+Σterms generated by the poles of H(s) |
| C. | Y(s)=k1/(s+jω)+(k1‘)/(s-jω)+Σterms generated by the poles of H(s) |
| D. | Y(s)=k1/(s-jω)+(k1‘)/(s-jω)+Σterms generated by the poles of H(s) |
| Answer» B. Y(s)=k1/(s+jω)+(k1‘)/(s+jω)+Σterms generated by the poles of H(s) | |
| 6. |
Let us assume x (t) = A cos(ωt + φ), what is the s-domain expression? |
| A. | Y(s)=H(s) A(Scos Ø-ω sinØ)/(S2-ω2) |
| B. | Y(s)=H(s) A(Scos Ø+ω sinØ)/(S2+ω2) |
| C. | Y(s)=H(s) A(Scos Ø-ω sinØ)/(S2+ω2) |
| D. | Y(s)=H(s) A(Scos Ø+ω sinØ)/(S2-ω2) |
| Answer» D. Y(s)=H(s) A(Scos Ø+ω sinØ)/(S2-ω2) | |
| 7. |
Let us assume x (t) = A cos(ωt + φ), then the Laplace transform of x (t) is? |
| A. | X(S)=A(Scos Ø-ω sinØ)/(S2-ω2) |
| B. | X(S)=A(Scos Ø+ω sinØ)/(S2+ω2) |
| C. | X(S)=A(Scos Ø+ω sinØ)/(S2-ω2) |
| D. | X(S)=A(Scos Ø-ω sinØ)/(S2+ω2) |
| Answer» E. | |
| 8. |
In the circuit shown below, if voltage across the capacitor is defined as the output signal of the circuit, then the transfer function is? |
| A. | H(s)=1/(S2 LC-RCS+1) |
| B. | H(s)=1/(S2 LC+RCS+1) |
| C. | H(s)=1/(S2 LC+RCS-1) |
| D. | H(s)=1/(S2 LC-RCS-1) |
| Answer» C. H(s)=1/(S2 LC+RCS-1) | |
| 9. |
In the circuit shown below, if current is defined as the response signal of the circuit, then determine the transfer function. |
| A. | H(s)=C/(S2 LC+RCS+1) |
| B. | H(s)=SC/(S2 LC-RCS+1) |
| C. | H(s)=SC/(S2 LC+RCS+1) |
| D. | H(s)=SC/(S2 LC+RCS-1) |
| Answer» D. H(s)=SC/(S2 LC+RCS-1) | |
| 10. |
THE_FINAL_STEADY_STATE_SOLUTION_FOR_Y_(T)_IN_THE_QUESTION_4_IS??$ |
| A. | A|H(jω) |cos⁡[ωt+Ø+ θ (ω)]. |
| B. | A|H(jω) |cos⁡[ωt-Ø+ θ (ω)]. |
| C. | A|H(jω) |cos⁡[ωt-Ø- θ (ω)]. |
| D. | A|H(jω) |cos⁡[ωt+Ø- θ (ω)]. |
| Answer» B. A|H(j‚âà√¨‚àö¬¢) |cos‚Äö√Ñ√∂‚àö√ñ¬¨‚àû[‚âà√¨‚àö¬¢t-‚Äö√†√∂‚àö‚â§+ ‚âà√≠‚Äö√†√® (‚âà√¨‚àö¬¢)]. | |
| 11. |
THE_VALUE_OF_K1_IN_THE_QUESTION_6_CONSIDERING_‚ÂÀ√≠‚ÄÖ√†√®_(‚ÂÀ√¨‚ÀÖ¬¢)_IS??$# |
| A. | |H(jω)|e<sup>j[θ (ω)+Ø]</sup> |
| B. | A/2|H(jω)|e<sup>j[θ (ω)+Ø]</sup> |
| C. | |H(jω)|e<sup>-j[θ (ω)+Ø]</sup> |
| D. | A/2 |H(jω) | e<sup>-j[θ (ω)+Ø]</sup> |
| Answer» C. |H(j‚âà√¨‚àö¬¢)|e<sup>-j[‚âà√≠‚Äö√†√® (‚âà√¨‚àö¬¢)+‚Äö√†√∂‚àö‚â§]</sup> | |
| 12. |
The relation between H (jω) and θ (ω) is?# |
| A. | H(jω)=e<sup>-jθ (ω)</sup> |
| B. | H(jω)=|H(jω)|e<sup>-jθ (ω)</sup> |
| C. | H(jω)=|H(jω)|e<sup>jθ (ω)</sup> |
| D. | H(jω)=e<sup>jθ (ω)</sup> |
| Answer» D. H(j‚âà√¨‚àö¬¢)=e<sup>j‚âà√≠‚Äö√†√® (‚âà√¨‚àö¬¢)</sup> | |
| 13. |
The value of k1 in the question 6 is? |
| A. | 1/2 H(jω)Ae<sup>jØ</sup> |
| B. | H(jω)Ae<sup>-jØ</sup> |
| C. | H(jω)Ae<sup>jØ</sup> |
| D. | 1/2 H(jω)Ae<sup>-jØ</sup> |
| Answer» E. | |
| 14. |
On taking the partial fractions for the response in the question 4, we get? |
| A. | Y(s)=k<sub>1</sub>/(s-jω)+(k<sub>1</sub><sup>‘</sup>)/(s+jω)+Σterms generated by the poles of H(s) |
| B. | Y(s)=k<sub>1</sub>/(s+jω)+(k<sub>1</sub><sup>‘</sup>)/(s+jω)+Σterms generated by the poles of H(s) |
| C. | Y(s)=k<sub>1</sub>/(s+jω)+(k<sub>1</sub><sup>‘</sup>)/(s-jω)+Σterms generated by the poles of H(s) |
| D. | Y(s)=k<sub>1</sub>/(s-jω)+(k<sub>1</sub><sup>‘</sup>)/(s-jω)+Σterms generated by the poles of H(s) |
| Answer» B. Y(s)=k<sub>1</sub>/(s+j‚âà√¨‚àö¬¢)+(k<sub>1</sub><sup>‚Äö√Ñ√∂‚àö√ë‚àö‚â§</sup>)/(s+j‚âà√¨‚àö¬¢)+‚âà√≠¬¨¬£terms generated by the poles of H(s) | |
| 15. |
The s-domain expression for the response for the input mentioned in question 4 is? |
| A. | Y(s)=H(s)A(Scos Ø-ω sinØ)/(S<sup>2</sup>-ω<sup>2</sup> ) |
| B. | Y(s)=H(s) A(Scos Ø+ω sinØ)/(S<sup>2</sup>+ω<sup>2</sup> ) |
| C. | Y(s)=H(s) A(Scos Ø-ω sinØ)/(S<sup>2</sup>+ω<sup>2</sup> ) |
| D. | Y(s)=H(s) A(Scos Ø+ω sinØ)/(S<sup>2</sup>-ω<sup>2</sup> ) |
| Answer» D. Y(s)=H(s) A(Scos ‚Äö√†√∂‚àö‚â§+‚âà√¨‚àö¬¢ sin‚Äö√†√∂‚àö‚â§)/(S<sup>2</sup>-‚âà√¨‚àö¬¢<sup>2</sup> ) | |
| 16. |
Let us assume x (t) = A cos(ωt + φ), then the Laplace transform of x (t) is?$ |
| A. | X(S)=A(Scos Ø-ω sinØ)/(S<sup>2</sup>-ω<sup>2</sup> ) |
| B. | X(S)=A(Scos Ø+ω sinØ)/(S<sup>2</sup>+ω<sup>2</sup> ) |
| C. | X(S)=A(Scos Ø+ω sinØ)/(S<sup>2</sup>-ω<sup>2</sup> ) |
| D. | X(S)=A(Scos Ø-ω sinØ)/(S<sup>2</sup>+ω<sup>2</sup> ) |
| Answer» E. | |
| 17. |
In the circuit shown in question 2, if voltage across the capacitor is defined as the output signal of the circuit, then the transfer function is? |
| A. | H(s)=1/(S<sup>2</sup> LC-RCS+1) |
| B. | H(s)=1/(S<sup>2</sup> LC+RCS+1) |
| C. | H(s)=1/(S<sup>2</sup> LC+RCS-1) |
| D. | H(s)=1/(S<sup>2</sup> LC-RCS-1) |
| Answer» C. H(s)=1/(S<sup>2</sup> LC+RCS-1) | |
| 18. |
The transfer function of a system having the input as X(s) and output as Y(s) is? |
| A. | Y(s)/X(s) |
| B. | Y(s) * X(s) |
| C. | Y(s) + X(s) |
| D. | Y(s) – X(s) |
| Answer» B. Y(s) * X(s) | |