Discussion
The given diagram is a section of transmission line with generating stations beyond either end of the generating station. The values of generated voltages are \(E_A = 430\) V, \(E_B = 330\) V. The values of impedances are \(Z_A = Z_B = Z_L = 50\ \Omega\). A relay (Rel) is connected as shown in the given figure.
The value of impedance seen by the relay is:
Answer: A
The two sources drive a current through the series path \(Z_A + Z_L + Z_B = 150\ \Omega\). With \(E_A\) and \(E_B\) in phase,
\[ I = \frac{430 - 330}{150} = 0.667\ \text{A} \]
The relay voltage is that of the bus after \(Z_A\):
\[ \begin{aligned} V = E_A - IZ_A = 430 - 0.667 \times 50 \\ &= 396.7\ \text{V} \end{aligned} \]
\[ \begin{aligned} Z_{relay} = \frac{V}{I} = \frac{396.7}{0.667} \\ &= 595\ \Omega \end{aligned} \]
Equivalently, \(Z = Z_L + Z_B + E_B/I = 100 + 495 = 595\ \Omega\). The answer is option A.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development