A three-bus, three-line system has uncompensated admittance $$Y=\begin{bmatrix}-j6&j3&j4\\j3&-j7&j5\\j4&j5&-j8\end{bmatrix}\ \mathrm{pu}.$$ After 50% compensation of all line shunt capacitance, the third diagonal element's imaginary contribution is
At bus 3, the two incident series branches contribute diagonal admittance −j(4+5)=−j9. The original diagonal −j8 therefore includes a +j1 capacitive shunt contribution.
Fifty-percent compensation removes half that capacitance contribution, leaving +j0.5. The new diagonal is −j9+j0.5=**−j8.5, B**. Strictly its imaginary part is −8.5; the choices retain the j factor as printed.
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