Question:medium
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:
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:
Show 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.
Updated On: Jul 20, 2026
- \(G^{2}+(B+0.5)^{2}\ge\dfrac{1}{4}\)
- \(B\ge1\)
- \((G-1)^{2}+B^{2}\le\dfrac{1}{2}\)
- \(G^{2}+(B-1)^{2}\ge\dfrac{1}{2}\)
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Think of the boundary as a set of points.
The line $X=1$ in the impedance plane is the set of all points $Z=R+j1$ as $R$ ranges over every real number. Each such point maps to one admittance point $Y=1/Z$.
Step 2: Convert a general point on the line.
\[ Y=\frac{1}{R+j}=\frac{R-j}{R^2+1} \]
so $G=\dfrac{R}{R^2+1}$ and $B=\dfrac{-1}{R^2+1}$.
Step 3: Eliminate the parameter $R$.
From the second equation, $R^2+1=-\dfrac{1}{B}$ (note $B$ is always negative here since $R^2+1>0$). Then
\[ G^2=\frac{R^2}{(R^2+1)^2}=\frac{R^2+1-1}{(R^2+1)^2}=\frac{1}{R^2+1}-\frac{1}{(R^2+1)^2} \]
Using $\frac{1}{R^2+1}=-B$:
\[ G^2=-B-B^2 \]
\[ G^2+B^2+B=0 \]
which is exactly the same circle equation as before, now confirmed by a completely different route, parametrizing by $R$ instead of substituting $Y=G+jB$ into $X=1$ directly.
Step 4: Complete the square.
\[ G^2+(B+0.5)^2=0.25 \]
This traces a circle centered at $(0,-0.5)$ of radius $0.5$ as $R$ sweeps from minus infinity to plus infinity; the point at $R=0$ gives $G=0,B=-1$, the bottom of the circle, and as $R$ grows large in either direction, $G,B$ both shrink toward zero, approaching the origin.
Step 5: Fix the inequality with one test point.
Pick a point clearly inside the trip zone, say $Z=j0.5$ (so $X=0.5\le1$). This gives $Y=1/(j0.5)=-j2$, so $G=0,B=-2$. Then $G^2+(B+0.5)^2=0+2.25=2.25$, which is greater than $0.25$. Since this point must sit inside the operate region, the operate condition is "greater than or equal to", the same conclusion as the direct method, now confirmed by two independent routes.
Step 6: State the final region.
\[ \boxed{G^{2}+(B+0.5)^{2}\ge\dfrac{1}{4}} \]
The line $X=1$ in the impedance plane is the set of all points $Z=R+j1$ as $R$ ranges over every real number. Each such point maps to one admittance point $Y=1/Z$.
Step 2: Convert a general point on the line.
\[ Y=\frac{1}{R+j}=\frac{R-j}{R^2+1} \]
so $G=\dfrac{R}{R^2+1}$ and $B=\dfrac{-1}{R^2+1}$.
Step 3: Eliminate the parameter $R$.
From the second equation, $R^2+1=-\dfrac{1}{B}$ (note $B$ is always negative here since $R^2+1>0$). Then
\[ G^2=\frac{R^2}{(R^2+1)^2}=\frac{R^2+1-1}{(R^2+1)^2}=\frac{1}{R^2+1}-\frac{1}{(R^2+1)^2} \]
Using $\frac{1}{R^2+1}=-B$:
\[ G^2=-B-B^2 \]
\[ G^2+B^2+B=0 \]
which is exactly the same circle equation as before, now confirmed by a completely different route, parametrizing by $R$ instead of substituting $Y=G+jB$ into $X=1$ directly.
Step 4: Complete the square.
\[ G^2+(B+0.5)^2=0.25 \]
This traces a circle centered at $(0,-0.5)$ of radius $0.5$ as $R$ sweeps from minus infinity to plus infinity; the point at $R=0$ gives $G=0,B=-1$, the bottom of the circle, and as $R$ grows large in either direction, $G,B$ both shrink toward zero, approaching the origin.
Step 5: Fix the inequality with one test point.
Pick a point clearly inside the trip zone, say $Z=j0.5$ (so $X=0.5\le1$). This gives $Y=1/(j0.5)=-j2$, so $G=0,B=-2$. Then $G^2+(B+0.5)^2=0+2.25=2.25$, which is greater than $0.25$. Since this point must sit inside the operate region, the operate condition is "greater than or equal to", the same conclusion as the direct method, now confirmed by two independent routes.
Step 6: State the final region.
\[ \boxed{G^{2}+(B+0.5)^{2}\ge\dfrac{1}{4}} \]
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