강의노트 3상 변압기

선간전압과 상전압의 관계

$\Delta$ 결선

$$V_{L}= V_{p}$$

$$I_{L}=\sqrt{3}I_{p}\angle -30^{\circ}$$

$$S =3S_p = 3V_p I_p = 3 V_L \dfrac{I_{L}}{\sqrt{3}} =\sqrt{3}V_{L}I_L$$

$$\begin{aligned} I_{A} &= I_{a}- I_{c} = I_{p}\angle 0 - I_{p}\angle -240 = I_{p}\left[1 -\left(-\dfrac{1}{2}+j\sqrt{\dfrac{3}{2}}\right)\right] \\ &= I_{p}\left(\dfrac{3}{2}-j\sqrt{\dfrac{3}{2}}\right) = I_{p}\sqrt{3}\angle -30 \end{aligned}$$

$$I_{B}= I_{p}\sqrt{3}\angle -150 \qquad I_{C}= I_{p}\sqrt{3}\angle -270$$

Y결선

$$V_{L}=\sqrt{3}V_{p}\angle 30^{\circ}$$

$$I_{L}= I_{p}$$

$$S =3S_p = 3V_p I_p = 3 \dfrac{V_{L}}{\sqrt{3}}I_L =\sqrt{3}V_{L}I_L$$

$$\begin{aligned} V_{ab} &= V_{an}+V_{nb} = V_{an}-V_{bn} = V_{p}\angle 0^{\circ}-V_{p}\angle -120^{\circ}\\ &=V_{p}\left(1-\left(-\dfrac{1}{2}-j\dfrac{\sqrt{3}}{2}\right)\right) =V_{p}\left(\dfrac{3}{2}+j\dfrac{\sqrt{3}}{2}\right) = V_{p}\sqrt{3}\angle 30 \end{aligned}$$

$$V_{bc}= V_{p}\sqrt{3}\angle -90 \qquad V_{ca}= V_{p}\sqrt{3}\angle -210$$