WYKAZAĆ, ŻE WATOMIERZ ELEKTRODYNAMICZNY MIERZY MOC CZYNNĄ:

$${\overset{\overline{}}{M}}_{N} = \ \frac{1}{T}\ \int_{0}^{T}{mn\ dt = \ }\frac{1}{T}\ \int_{0}^{T}{i_{1\ }\ i_{2}\ \frac{\text{dM}_{12}}{d \propto}dt = \ \frac{1}{T}}\int_{0}^{T}{\frac{U_{m}}{R}\sin\text{ωt}I_{m}\sin\left\lbrack \omega t + \sphericalangle\ \left( U;I \right) \right\rbrack\frac{\text{dM}_{12}}{d \propto}dt = \ \frac{1}{T}}\frac{U_{m}I_{m}}{R}\ \frac{\text{dM}_{12}}{d \propto}\int_{0}^{T}{\sin\text{ωt}\sin\left\lbrack \omega t + \sphericalangle\ \left( U;I \right) \right\rbrack dt = \ \frac{1}{T}\ }\ \frac{2UI}{R}\ \frac{\text{dM}_{12}}{d \propto}\int_{0}^{T}{\operatorname{}\left\lbrack cos\sphericalangle\ \left( U;I \right) \right\rbrack - \cos\left\lbrack \ 2\omega t + \sphericalangle\ \left( U;I \right)\text{dt\ } \right\rbrack = \ \frac{1}{\text{TR}}\ }\text{UI}\frac{\text{dM}_{12}}{d \propto}(\int_{0}^{T}\begin{matrix} cos\sphericalangle\ \left( U;I \right)\ dt + \int_{0}^{T}{\cos\left\lbrack 2\omega t + \sphericalangle\ \left( U;I \right)\text{dt} \right\rbrack} \\ \ \\ \end{matrix}$$

$${\overset{\overline{}}{M}}_{N} = \ \frac{1}{R}\ \frac{\text{dM}_{12}}{d \propto}\text{\ UJ}\cos\sphericalangle\ \left( U,J \right)$$

Zatem

$$\cos{\sphericalangle\ \left( U,I \right) = \cos{\varphi,}}\ {\overset{\overline{}}{M}}_{N} = \ M_{2}$$

$$\alpha = \ \frac{1}{k_{z}}\ \frac{1}{R}\ \frac{\text{dM}_{12}}{d \propto}\text{\ UJcosφ}$$

$P = kR\frac{\text{dM}_{12}}{d \propto}*\alpha$ co wynika, że P = C_w * α_w