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I was trying to find the relationship between these two online, but I found out two completely opposite answers, and I'm wondering which one is the right one.

Is B = -1/X or is B = 1/X ?

Edward B
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    I've always seen it as B=1/X, though I'm not confident enough in that to say so definitively. Susceptance doesn't come up very often in my field of study. – Hearth May 06 '18 at 14:15

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Impedance consists of a real part (reistance) and an imaginery part (reactance). Admittance, defined as the reciprocal of impedance, also has a real part (conduction) and an imaginery part (susceptance). If you have a pure reactance, for which the resistance is 0, then the formula reduces to admittance = 1/jX, where X is the reactance. This, in return, simplifies to admittance = -1/X. Thus the answer is that B, the susceptance is equal to -1/X.

Barry
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Susceptance (\$B\$), is a component of Admittance (\$Y\$). Admittance (\$Y\$) is defined as the current (\$I\$) flowing per unit of applied voltage (\$V\$). \$Y= I/V\$ which is the reciprocal of Impedance (\$Z\$), where \$Z= V/I\$. Impedance (\$Z\$) is a complex number having a real Resistance (\$R\$) and an imaginary component Reactance (\$X\$), such that \$Z=R+jX\$.

But reactance itself is made up of inductive or capacitive components \$Z=R+j(X_L-X_C)\$. THIS is where that elusive (-) comes from. So it is that Admittance (\$Y\$) is a complex number comprising real Conductance (\$G\$) and imaginary Susceptance (\$B\$).

\$Y=G +jB\$ which in more detail is \$Y=G+j(B_L-B_C)\$. Now if there is a pure inductance then \$G\$ and \$B_C\$ are both zero and \$Y = jB\$ if there is a pure capacitance then \$Y = -jB\$. The positive and negative indicate whether the admittance relates to pure inductance or capacitance.

Adam Haun
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derek macfarlane
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Reactance is positive = X (A reactance would be 2000 Ohms not -2000 Ohms)

So susceptance is also positive = 1/X

Capacitor's impedance = -jXC

Inductor's impedance = jXL