電験三種の「電力」で重要な公式一覧！問題を解くのに必要な公式をご紹介

$P_o=9.8QH$

$P_o=9.8Q\left(H_a–h_g\right)$

$P_g=9.8QH{\eta }_t{\eta }_g$

$P_g=9.8Q\left(H_a–h_g\right){\eta }_t{\eta }_g$

$P_m={\displaystyle \frac{9.8QH}{{\eta }_m{\eta }_p}}$

$P_m={\displaystyle \frac{9.8Q\left(H_a+h_p\right)}{{\eta }_m{\eta }_p}}$

$P_S=P_G–P_L$

$P_S=P_G\left(1–L\right)$

$E=\Delta m{c}^2$

$E=9\times {10}^{-4}m{c}^2$

$P={\displaystyle \frac{1}{2}}m{v}^2$

$P={\displaystyle \frac{1}{2}}\rho A{v}^3$

$\%Z={\displaystyle \frac{I_{n}Z}{E_n}}\times 100$

$\%Z={\displaystyle \frac{P_{n}Z}{{E_n}^2}}\times 100$

$\%Z={\displaystyle \frac{\sqrt{3}{I}_{n}Z}{V_n}}\times 100$

$\%Z={\displaystyle \frac{P_{n}Z}{{V_n}^2}}\times 100$

$P_a={\displaystyle \frac{\%Z_b{P}_{\mathrm{an}}}{\%Z_a{P}_{\mathrm{bn}}+\%Z_b{P}_{\mathrm{an}}}}\times \mathrm{P}$

$P_b={\displaystyle \frac{\%Z_a{P}_{\mathrm{bn}}}{\%Z_a{P}_{\mathrm{bn}}+\%Z_b{P}_{\mathrm{an}}}}\times \mathrm{P}$

$\cos \theta ={\displaystyle \frac{P}{S}}$

$\cos \theta ={\displaystyle \frac{R}{Z}}$

$V_d=V_s–V_r$

$V_d=\sqrt{3}I\left(R\cos \theta +X\sin \theta \right)$

$V_d={\displaystyle \frac{RP+XQ}{V_r}}$

$P=\omega C{V}^2\tan \delta$

$P=2\pi fC{V}^2\tan \delta$

$I={\displaystyle \frac{\omega CV}{\sqrt{3}}}$

$I={\displaystyle \frac{2\pi fCV}{\sqrt{3}}}$

$Q=\sqrt{3}VI$

$Q=2\pi fC{V}^2$

$P=3V_{\text{相}}{I}_{\text{相}}\cos \theta$

$P=\sqrt{3}{V}_{\text{線}}{I}_{\text{線}}\cos \theta$

$P=3V_{\text{相}}{I}_{\text{相}}\cos \theta$

$P=\sqrt{3}{V}_{\text{線}}{I}_{\text{線}}\cos \theta$

$I=3\omega CE$

$I=\sqrt{3}\omega CV$