A three-phase balanced voltage is applied to the load shown. The phase sequence is RYB. The ratio $\frac{\left|I_B\right|}{\left|I_R\right|}$ is  ______.

The correct answer is 1.

1. Circuit Analysis and Identification of Node Voltages:
From the given circuit diagram:
- The R-phase terminal is connected to the common junction node (let's call its voltage $V_N$) through a capacitor of impedance $Z_C=-j10\Omega$.
- The B-phase terminal is connected directly to the junction node without any impedance. Therefore, the node voltage is:

$V_N=V_B$

- The Y-phase terminal is connected to the junction node through an inductor of impedance $Z_L=j10\Omega$.

2. Expressing Phase Currents:
We can express the line currents entering the load using Ohm's Law:

$I_R=\frac{V_R-V_B}{-j10}=\frac{V_{\text{RB}}}{-j10}$

$I_Y=\frac{V_Y-V_B}{j10}=\frac{V_{\text{YB}}}{j10}$

Applying Kirchhoff's Current Law (KCL) at the junction node, the current $I_B$ is:

$I_B=-(I_R+I_Y)$

3. Defining the Balanced 3-Phase Voltages (RYB Sequence):
Let $V_p$ be the RMS phase voltage. With the RYB sequence, we define the phase voltages as:
$V_R=V_p\angle 0^{\circ }$
$V_Y=V_p\angle -{120}^{\circ }$
$V_B=V_p\angle {120}^{\circ }$

Let's find the line-to-line voltages $V_{\text{RB}}$ and $V_{\text{YB}}$:

$V_{\text{RB}}=V_R-V_B=V_p(1-(-0.5+j\frac{\sqrt{3}}{2}\left)\right)=V_p(1.5-j\frac{\sqrt{3}}{2})=\sqrt{3}{V}_p\angle -{30}^{\circ }$

$V_{\text{YB}}=V_Y-V_B=V_p(-0.5-j\frac{\sqrt{3}}{2}-(-0.5+j\frac{\sqrt{3}}{2}\left)\right)=-j\sqrt{3}{V}_p=\sqrt{3}{V}_p\angle -{90}^{\circ }$

4. Calculating the Currents:
Substituting these values back into the expressions for $I_R$ and $I_Y$:

$I_R=\frac{\sqrt{3}{V}_p\angle -{30}^{\circ }}{10\angle -{90}^{\circ }}=\frac{\sqrt{3}{V}_p}{10}\angle {60}^{\circ }$

$I_Y=\frac{\sqrt{3}{V}_p\angle -{90}^{\circ }}{10\angle {90}^{\circ }}=\frac{\sqrt{3}{V}_p}{10}\angle -{180}^{\circ }$

Let $I_0=\frac{\sqrt{3}{V}_p}{10}$. Thus:
$I_R=I_0\angle {60}^{\circ }=I_0(0.5+j\frac{\sqrt{3}}{2})$
$I_Y=I_0\angle -{180}^{\circ }=-I_0$

Now compute $I_B$:

$I_B=-(I_R+I_Y)=-I_0(0.5+j\frac{\sqrt{3}}{2}-1)=-I_0(-0.5+j\frac{\sqrt{3}}{2})=I_0(0.5-j\frac{\sqrt{3}}{2})=I_0\angle -{60}^{\circ }$

Comparing the magnitudes:
$\left|I_R\right|=I_0$
$\left|I_B\right|=I_0$

5. Final Ratio:
The ratio of the magnitude of the B-phase line current to the R-phase line current is:

$\frac{\left|I_B\right|}{\left|I_R\right|}=\frac{I_0}{I_0}=1$