Question

The operating characteristic of a reactance relay is given by \(X\le1\ \Omega\), where \(X\) is the reactance calculated by the relay. Its operating characteristic in the admittance plane (\(G\)-\(B\) plane, where \(G\) and \(B\) denote conductance and susceptance, respectively, expressed in \(\Omega^{-1}\)) is given by:

• A. \(G^{2}+(B+0.5)^{2}\ge\dfrac{1}{4}\)
• B. \(B\ge1\)
• C. \((G-1)^{2}+B^{2}\le\dfrac{1}{2}\)
• D. \(G^{2}+(B-1)^{2}\ge\dfrac{1}{2}\)

Hint:

Substitute R=G/(G-squared+B-squared) and X=-B/(G-squared+B-squared) into X=1, complete the square in B, then test one known operating point to fix the inequality direction.

Answer 1

The Correct Option is A

Step 1: Write the impedance in terms of admittance.
Let \(Z=R+jX\) be the impedance measured by the relay, and let \(Y=G+jB=1/Z\) be the admittance. Rationalizing,
\[ Z=\frac{1}{Y}=\frac{1}{G+jB}=\frac{G-jB}{G^2+B^2} \]

Step 2: Read off R and X.
Comparing real and imaginary parts,
\[ R=\frac{G}{G^2+B^2},\qquad X=\frac{-B}{G^2+B^2} \]

Step 3: Apply the boundary of the reactance characteristic.
The relay's boundary line is \(X=1\). Substituting,
\[ \frac{-B}{G^2+B^2}=1 \]
\[ -B=G^2+B^2 \]
\[ G^2+B^2+B=0 \]

Step 4: Complete the square in \(B\).
\[ G^2+\left(B+\frac{1}{2}\right)^2=\frac{1}{4} \]
This is a circle in the \(G\)-\(B\) plane centered at \((0,-0.5)\) with radius \(0.5\).

Step 5: Decide which side of the circle is the operate region.
Take a point deep inside the trip region in the impedance plane, say \(R=0,\ X=0.5\), which satisfies \(X\le1\). Its admittance is
\[ Y=\frac{1}{j0.5}=-j2\quad\Rightarrow\quad G=0,\ B=-2 \]
Check:
\[ G^2+(B+0.5)^2=0+(-1.5)^2=2.25\ge\frac{1}{4} \]
So points that operate the relay satisfy the "greater than or equal to" inequality, meaning they lie on or outside this circle.

Step 6: Double-check with a non-operating point.
Take \(R=0,\ X=2\), outside the trip zone since \(X>1\). Then \(Y=1/(j2)=-j0.5\), so \(G=0,\ B=-0.5\).
\[ G^2+(B+0.5)^2=0+0^2=0<\frac{1}{4} \]
This point fails the inequality, confirming that the operate region is indeed \(G^2+(B+0.5)^2\ge\dfrac{1}{4}\).

Step 7: Analyze the options.

(A) \(G^2+(B+0.5)^2\ge1/4\): Matches the derived boundary and direction exactly. Correct.

(B) \(B\ge1\): A straight line condition, not a circle; it does not come from transforming \(X=1\). Incorrect.

(C) \((G-1)^2+B^2\le1/2\): Wrong center, wrong radius, and wrong side. Incorrect.

(D) \(G^2+(B-1)^2\ge1/2\): Wrong center, the sign of the \(B\) shift is flipped, and wrong radius. Incorrect.

Final Answer:
\[ \boxed{G^{2}+(B+0.5)^{2}\ge\frac{1}{4}} \]