We are given a system with three alternating voltage sources forming a balanced 3-phase set:

• $v_1 = V \sin(\omega t) \implies \mathbf{V}_1 = V \angle 0^\circ$
• $v_2 = V \sin(\omega t + 120^\circ) \implies \mathbf{V}_2 = V \angle 120^\circ$
• $v_3 = V \sin(\omega t + 240^\circ) \implies \mathbf{V}_3 = V \angle 240^\circ$

The components connected to these sources have the following reactances:

• Resistor: $Z_1 = R$
• Capacitor: The reactance modulus is equal to $R$, so $X_C = R$. In phasor notation, capacitive impedance is $Z_2 = -jX_C = -jR$.
• Inductor: The reactance modulus is equal to $R$, so $X_L = R$. In phasor notation, inductive impedance is $Z_3 = jX_L = jR$.

Let the potential at the common neutral point (left side) be $0$ and the potential at junction P be $\mathbf{V}_P$. According to Kirchhoff’s Current Law (KCL), the sum of currents leaving junction P is zero:

$$ \frac{\mathbf{V}_P – \mathbf{V}_1}{Z_1} + \frac{\mathbf{V}_P – \mathbf{V}_2}{Z_2} + \frac{\mathbf{V}_P – \mathbf{V}_3}{Z_3} = 0 $$

Substituting the impedance values:

$$ \frac{\mathbf{V}_P – \mathbf{V}_1}{R} + \frac{\mathbf{V}_P – \mathbf{V}_2}{-jR} + \frac{\mathbf{V}_P – \mathbf{V}_3}{jR} = 0 $$

Multiplying the entire equation by $R$ to eliminate denominators:

$$ (\mathbf{V}_P – \mathbf{V}_1) + \frac{\mathbf{V}_P – \mathbf{V}_2}{-j} + \frac{\mathbf{V}_P – \mathbf{V}_3}{j} = 0 $$

Recall that $\frac{1}{j} = -j$ and $\frac{1}{-j} = j$. The equation becomes:

$$ (\mathbf{V}_P – \mathbf{V}_1) + j(\mathbf{V}_P – \mathbf{V}_2) – j(\mathbf{V}_P – \mathbf{V}_3) = 0 $$

Expanding the terms:

$$ \mathbf{V}_P – \mathbf{V}_1 + j\mathbf{V}_P – j\mathbf{V}_2 – j\mathbf{V}_P + j\mathbf{V}_3 = 0 $$

Notice that $+j\mathbf{V}_P$ and $-j\mathbf{V}_P$ cancel each other out:

$$ \mathbf{V}_P – \mathbf{V}_1 – j\mathbf{V}_2 + j\mathbf{V}_3 = 0 $$ $$ \mathbf{V}_P = \mathbf{V}_1 + j(\mathbf{V}_2 – \mathbf{V}_3) $$

Now we substitute the phasor values of the sources. Using standard polar-to-rectangular conversion ($V \angle \theta = V(\cos\theta + j\sin\theta)$):

• $\mathbf{V}_1 = V$
• $\mathbf{V}_2 = V(-\frac{1}{2} + j\frac{\sqrt{3}}{2})$
• $\mathbf{V}_3 = V(-\frac{1}{2} – j\frac{\sqrt{3}}{2})$

First, let’s calculate the term $(\mathbf{V}_2 – \mathbf{V}_3)$:

$$ \mathbf{V}_2 – \mathbf{V}_3 = \left[ V\left(-\frac{1}{2} + j\frac{\sqrt{3}}{2}\right) \right] – \left[ V\left(-\frac{1}{2} – j\frac{\sqrt{3}}{2}\right) \right] $$ $$ \mathbf{V}_2 – \mathbf{V}_3 = V\left( j\frac{\sqrt{3}}{2} – (-j\frac{\sqrt{3}}{2}) \right) = V(j\sqrt{3}) $$

Now substitute this back into the expression for $\mathbf{V}_P$:

$$ \mathbf{V}_P = \mathbf{V}_1 + j(j\sqrt{3}V) $$ $$ \mathbf{V}_P = V + j^2\sqrt{3}V $$

Since $j^2 = -1$:

$$ \mathbf{V}_P = V – \sqrt{3}V = V(1 – \sqrt{3}) $$

The phasor result is:

$$ \mathbf{V}_P = V(1 – \sqrt{3}) $$

Since $\sqrt{3} \approx 1.732$, the term $(1 – \sqrt{3})$ is a negative real number. We can rewrite this to express the magnitude and phase explicitly:

$$ \mathbf{V}_P = -(\sqrt{3} – 1)V $$

A negative sign in phasors corresponds to a phase shift of $180^\circ$.

$$ \text{Magnitude} = (\sqrt{3} – 1)V $$ $$ \text{Phase Angle} = 180^\circ $$

Converting this back to the time domain: