We suppose that all the multi-phased electrical systems are properly balanced: $\forall n \in \{1,2\}; Iph_{n} = Iph_{n+1}$.

Let's call $A = (p_{a}, V_{a}, I_{a}, Iph_{a})$ and $B = (p_{b}, V_{b}, I_{b}, Iph_{b})$ two multi-phased electrical systems where $p_{a},p_{b} \in \mathbb{N}$ are respectively the number of phases, $V_{a}, V_{b} \in \mathbb{N}$ are the voltage per phase, $I_{a}, I_{b} \in \mathbb{N}$ are the total intensity and $Iph_{a}, Iph_{b} \in \mathbb{N}$ the intensity per phase.
The intensities properties can be deduced one from the other: $I_{a} = p_{a} \times Iph_{a}$ and $I_{b} = p_{b} \times Iph_{b}$.

• $A$ and $B$ are interconnected serially : $A \triangleleft \triangleright B$.

How to evaluate the properties of the resulting system $A \triangleleft \triangleright B$?
$A \triangleleft \triangleright B = (p = min(p_{a},p_{b}), V_{A \triangleleft \triangleright B} = min(V_{a}, V_{b}), I_{A \triangleleft \triangleright B} = min(I_{a}, I_{b}), Iph_{A \triangleleft \triangleright B} = min(Iph_{a},Iph_{b}))$

Let's say you only have $I_{a}, I_{b}$ in $A$ and $B$. We can deduce $Iph_{a}, Iph_{b}$ from them.
How can we deduce $Iph_{A \triangleleft \triangleright B}$?
$Iph_{A \triangleleft \triangleright B} = min(I_{a}, I_{b}) \div min(p_{a},p_{b})$
Division evaluation:

• $I_{a} < I_{b} \land p_{a} < p_{b} \Rightarrow I_{a} \ mod \ p_{a} = 0 \land Iph_{A \triangleleft \triangleright B} \in \mathbb{N}$ ;
• $I_{a} > I_{b} \land p_{a} < p_{b} \land p_{a}, p_{b} \in \{2,3\} \land I_{b} \ mod \ p_{a} = 0 \Rightarrow Iph_{A \triangleleft \triangleright B} \in \mathbb{N}$ ;
• $I_{a} > I_{b} \land p_{a} > p_{b} \Rightarrow I_{b} \ mod \ p_{b} = 0 \land Iph_{A \triangleleft \triangleright B} \in \mathbb{N}$;
• $I_{a} > I_{b} \land p_{a} < p_{b} \land p_{a}, p_{b} \in \{2,3\} \land I_{b} \ mod \ p_{a} = 0 \Rightarrow Iph_{A \triangleleft \triangleright B} \in \mathbb{N}$ ;
• $I_{a} > I_{b} \land p_{a} < p_{b} \land p_{a}, p_{b} \in \{2,3\} \land I_{b} \ mod \ p_{a} \neq 0 \Rightarrow Iph_{A \triangleleft \triangleright B} \notin \mathbb{N}$ ;
• $I_{a} < I_{b} \land p_{a} > p_{b} \land p_{a}, p_{b} \in \{2,3\} \land I_{a} \ mod \ p_{b} \ne 0 \Rightarrow Iph_{A \triangleleft \triangleright B} \notin \mathbb{N}$ ;

We can expose at least two cases where $Iph_{A \triangleleft \triangleright B} \notin \mathbb{N}$ if $p_{a}, p_{b} \in \{2,3\}$

Let's say you only have $Iph_{a}, Iph_{b}$ in $A$ and $B$. We can deduce $I_{a}, I_{b}$ from them.
How can we deduce $I_{A \triangleleft \triangleright B}$?
$I_{A \triangleleft \triangleright B} = min(p_{a},p_{b}) \times min(I_{a}, I_{b}) \in \mathbb{N}$

So if you model an AC electrical system with the properties $(p, V, Iph)$, you can deduce the properties of any interconnexion of them within the natural integer set.

$I$ is the line current.
$V$ is the line-to-neutral voltage.
$U$ is the line-to-line voltage: $U = \sqrt{3} \times V$.

$$I = p \times Iph \\ Pph = Vph \times Iph \times \cos(\varphi) \\ P = \sum_{n=1}^{p} Pph_{n} = \sum_{n=1}^{p} Vph_{n} \times Iph_{n} \times \cos(\varphi) = p \times Vph \times Iph \times \cos(\varphi) = Vph \times I \times \cos(\varphi) \\ Qph = Vph \times Iph \times \sin(\varphi) \\ Q = \sum_{n=1}^{p} Qph_{n} = \sum_{n=1}^{p} Vph_{n} \times Iph_{n} \times \sin(\varphi) = p \times Vph \times Iph \times \sin(\varphi) = Vph \times I \times \sin(\varphi) \\ Sph = Vph \times Iph \\ S = \sum_{n=1}^{p} Sph_{n} = \sum_{n=1}^{p} Vph_{n} \times Iph_{n} = p \times Vph \times Iph = Vph \times I$$

$J$ is the phase current.
$I$ is the line current: $I = \sqrt{3} \times J$.
$U$ is the line-to-line voltage.

$$J = p \times Jph \\ Pph = Uph \times Jph \times \cos(\varphi) \\ P = \sum_{n=1}^{p} Pph_{n} = \sum_{n=1}^{p} Uph_{n} \times Jph_{n} \times \cos(\varphi) = p \times Uph \times Jph \times \cos(\varphi) = Uph \times J \times \cos(\varphi) \\ Qph = Uph \times Jph \times \sin(\varphi) \\ Q = \sum_{n=1}^{p} Qph_{n} = \sum_{n=1}^{p} Uph_{n} \times Jph_{n} \times \sin(\varphi) = p \times Uph \times Jph \times \sin(\varphi) = Uph \times J \times \sin(\varphi) \\ Sph = Uph \times Jph \\ S = \sum_{n=1}^{p} Sph_{n} = \sum_{n=1}^{p} Uph_{n} \times Jph_{n} = p \times Uph \times Jph = Uph \times J$$

Low level class of electrical helper methods implementation:

/**
 * Targeted to AC related values calculation.
 */
export class ACElectricUtils {
  static amperageTotal(nbOfPhases: number, Iph: number): number {
    return nbOfPhases * Iph;
  }
 
  static powerPerPhase(V: number, Iph: number, cosPhi = 1): number {
    const powerPerPhase = V * Iph * cosPhi;
    if (cosPhi === 1) {
      return powerPerPhase;
    }
    return Math.round(powerPerPhase);
  }
 
  static powerTotal(nbOfPhases: number, V: number, Iph: number, cosPhi = 1): number {
    return nbOfPhases * ACElectricUtils.powerPerPhase(V, Iph, cosPhi);
  }
 
  static amperageTotalFromPower(P: number, V: number, cosPhi = 1): number {
    const amperage = P / (V * cosPhi);
    if (cosPhi === 1 && P % V === 0) {
      return amperage;
    }
    return Math.round(amperage);
  }
 
  static amperagePerPhaseFromPower(nbOfPhases: number, P: number, V: number, cosPhi = 1): number {
    const amperage = ACElectricUtils.amperageTotalFromPower(P, V, cosPhi);
    const amperagePerPhase = amperage / nbOfPhases;
    if (amperage % nbOfPhases === 0) {
      return amperagePerPhase;
    }
    return Math.round(amperagePerPhase);
  }
}
 
/**
 * Targeted to DC related values calculation.
 */
export class DCElectricUtils {
  static power(V: number, I: number): number {
    return V * I;
  }
 
  static amperage(P: number, V: number): number {
    const amperage = P / V;
    if (P % V === 0) {
      return amperage;
    }
    return Math.round(amperage);
  }
}

AC electrical component interface:

interface ElectricACComponent {
  nPhases: number;
  Iph: number;
  V: number;
  I?: number;
  P?: number;
}

Electrical properties calculation of an electric car connected to a AC charging station implementation (the electrical properties of the connexion cable are always beyond):

const Car = {
  nPhases: 2,
  Iph: 18,
  V: 220,
};
 
const ChargingStation = {
  nPhases: 3,
  Iph: 32,
  V: 230,
};
 
function getCombinedElectricObject(Car, ChargingStation): ElectricACComponent {
  return {
    nPhases: Math.min(Car.nPhases, ChargingStation.nPhases),
    Iph: Math.min(Car.Iph, ChargingStation.Iph),
    V: Math.min(Car.V, ChargingStation.V),
    I: ACElectricUtils.amperageTotal(this.nPhases, this.Iph),
    P: ACElectricUtils.powerTotal(this.nPhases, this.Iph, this.V),
  };
}

And now for example you can define a fine grained OCPP charging profile with the following schedule for the electric car connected to the charging station that will include the number of phases used and the max intensity per phase:

const ChargingSchedulePeriod = {
  startPeriod: 60,
  limit: 15, // Sanity check: ensure the limit is below getCombinedElectricObject(Car, ChargingStation).Iph
  numberPhases: getCombinedElectricObject(Car, ChargingStation).nPhases,
};

See first section.

See first section.

Composants symétriques : https://fr.qwe.wiki/wiki/Symmetrical_components