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Byju's Answer
Standard XII
Physics
Combination of Inductors
For the 4-bus...
Question

For the 4-bus system, containing transmission lines and transformers. The admittance matrix is Ybus=⎡⎢ ⎢ ⎢ ⎢⎣−j20j2000j20−j60j4000j40−j60j2000j20−j25⎤⎥ ⎥ ⎥ ⎥⎦

If a transmission line having π-equivalent circuit as shown below is connected between 2nd and 4th buses. Then the modified Ybus matrix will be


Where, Zse = series impedance of transmission line.

Ysh= shunt admittance of transmission line.

A
Ybus=⎡⎢ ⎢ ⎢ ⎢⎣−j20j2000j20−j50j4000j40−j60j2000j20−j15⎤⎥ ⎥ ⎥ ⎥⎦
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B
Ybus=⎡⎢ ⎢ ⎢ ⎢⎣−j20j2000j20−j50j4000j40−j60j200−j20j20−j15⎤⎥ ⎥ ⎥ ⎥⎦
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C
Ybus=⎡⎢ ⎢ ⎢ ⎢⎣−j20j2000j20−j50j40−j200j40−j60j200−j20j20−j15⎤⎥ ⎥ ⎥ ⎥⎦
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D
Ybus=⎡⎢ ⎢ ⎢ ⎢⎣−j20j2000j20−j40j40−j200j40−j60j200−j20j20−j5⎤⎥ ⎥ ⎥ ⎥⎦
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Solution

The correct option is C Ybus=⎡⎢ ⎢ ⎢ ⎢⎣−j20j2000j20−j50j40−j200j40−j60j200−j20j20−j15⎤⎥ ⎥ ⎥ ⎥⎦
Only Y22,Y24,Y42,Y44 will change because transmission line is connected between 2nd and 4th buses.

Y22=−j60+1Zse+Ysh2
=−j60+1j0.1+j20=−j60−j10+j20=−j50


Y24=Y42=0−Ysh2=−j20

Y44=−j25+1Zse+Ysh2=−j25+1j0.1+j20=−j25−j10+j20

Y44=−j15

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