# Instantaneous Reactive Power Theory in the Geometric Algebra Framework
## 2 Instantaneous Reactive Power Theory in Three-Phase Systems: A New Approach
### Instantaneous Power/Instantaneous Reactive Currents  
In this section, from the power terms defined in the previous section, the current components are determined. First, the inverse of the vector \(\mathbf{u}\) is defined. **Definition 3**. : _The inverse of the voltage vector, \(\mathbf{u}^{-1}\), is determined from the relation (16):_  
\[\mathbf{u}^{-1}=\frac{\mathbf{u}^{\dagger}}{\mathbf{u}^{\dagger}\mathbf{u}}= \frac{\mathbf{u}}{\mathbf{u}^{\dagger}\mathbf{u}} \tag{16}\]  
_and, therefore, (17) is verified:_  
\[\mathbf{u}\mathbf{u}^{-1}=\mathbf{u}^{-1}\mathbf{u}=1 \tag{17}\]The denominator of (16) identifies the square of the voltage vector norm:  
\[\mathbf{u}^{\intercal}\mathbf{u}\equiv\|\mathbf{u}\|^{2}=u_{1}^{2}+u_{2}^{2}+u_{ 3}^{2}=u^{2} \tag{18}\]  
In the framework of the IRPT, expression (18) represents the square of the instantaneous norm of the voltage vector [1]. **Theorem 2**. : _The current vector \(\mathbf{i}\) of a three-phase system with voltage \(\mathbf{u}\) is determined from the instantaneous power multivector according to expression (19):_  
\[\mathbf{i}=\frac{\mathbf{u}^{\intercal}\mathbf{s}}{\mathbf{u}^{\intercal} \mathbf{u}}=\frac{\mathbf{u}\mathbf{s}}{\mathbf{u}^{\intercal}\mathbf{u}} \tag{19}\]  
In fact, by multiplying the left side of (3) by the inverse of the voltage vector, it follows that:  
\[\mathbf{u}^{-\intercal}\mathbf{u}\mathbf{i}=\mathbf{u}^{-\intercal}\mathbf{s} \tag{20}\]  
since by application of the associative property of the geometric product in (20), [11], and the relation (17):  
\[\mathbf{i}=\mathbf{u}^{-\intercal}\mathbf{s}=\frac{\mathbf{u}^{\intercal} \mathbf{s}}{\mathbf{u}^{\intercal}\mathbf{u}}=\frac{\mathbf{u}\mathbf{s}}{ \left\|\mathbf{u}\right\|^{2}}=\frac{\mathbf{u}\mathbf{s}}{u^{2}} \tag{21}\]  
also taking into account that, for the 1-vectors, from (9), it is true that:  
\[\mathbf{u}^{\intercal}=\mathbf{u} \tag{22}\]  
**Theorem 3**. : _The current \(\mathbf{i}\) decomposes into two terms, the instantaneous power current \(\mathbf{i_{p}}\), and the instantaneous reactive current, \(\mathbf{i_{q}}\):_  
\[\mathbf{i}=\mathbf{i_{p}}+\mathbf{i_{q}}=\frac{\mathbf{u}p}{\left\|\mathbf{u} \right\|^{2}}+\frac{\mathbf{u}\mathbf{q}}{\left\|\mathbf{u}\right\|^{2}} \tag{23}\]  
_Furthermore, both components are orthogonal and verify the relation (24):_  
\[\left\|\mathbf{i}\right\|^{2}=i^{2}=\left\|\mathbf{i_{p}}\right\|^{2}+\left\| \mathbf{i_{q}}\right\|^{2}=\hat{\ell}_{p}^{2}+\hat{\ell}_{q}^{2} \tag{24}\]  
In fact, from (21) it follows that:  
\[\mathbf{i}=\frac{\mathbf{u}}{u^{2}}(\langle\mathbf{s}\rangle_{0}+\langle \mathbf{s}\rangle_{2})=\frac{\mathbf{u}}{u^{2}}(p+\mathbf{q}) \tag{25}\]  
The first term of (25) is the instantaneous power current, \(\mathbf{i_{p}}\):  
\[\mathbf{u}\mathbf{u}^{-1}=\mathbf{u}^{-1}\mathbf{u}=1 \tag{17}\]0  
The second term is the instantaneous reactive current, \(\mathbf{i_{q}}\):  
\[\mathbf{u}\mathbf{u}^{-1}=\mathbf{u}^{-1}\mathbf{u}=1 \tag{17}\]1  
Both current terms are orthogonal, since:  
\[ \begin{array}{ll}\mathbf{i}^{\intercal}\mathbf{p}&=\frac{\mathbf{u}}{u^{2}} p\\ &=\frac{p}{u^{2}}(u_{1}e_{1}+u_{2}e_{2}+u_{3}e_{3})\end{array}