When calculating the active power in the 3 phase Star, and Delta formations, there are two different currents used depending on the layout. I have shown my working below, with what I have been taught.

However, I am wondering why when calculating the Active power for the star connected formation we use \$I_{phase}\$ for the calculation. Whereas in the Delta connected formation we use \$I_{line}\$. What is the reason for this?

Solution sheet


2 Answers 2


You are using \$I_{LINE}\$ in both.

There is no dot (junction), so the current that flows in phase must flow in transmission line. So in a star or wye, \$I_{LINE}\ =\ I_{PHASE}\$ (and \$V_{LINE}\ =\ \sqrt {3}\ V_{PHASE}\$). You show this in your answer.

In a delta, \$I_{LINE}\ =\ \sqrt {3}\ I_{PHASE}\$ (and \$V_{LINE}\ =\ V_{PHASE}\$). A component of two phase currents make up the line current. There is a dot (junction).

For line quantities: $$P_T = \sqrt {3}\ V_{LINE}\ I_{LINE}\ cos\ θ $$ For phase quantities: $$P_T = 3\ V_{PHASE}\ I_{PHASE}\ cos\ θ $$

So in your first answer, \$V_{L-L}\$ which is line voltage \$V_{LINE}\ =\ 415V\$, which means a \$V_{PHASE}\ = \frac {415V} {\sqrt {3}}\ =\ 239.6V\$.

$$ I_{PHASE}\ =\ \frac {V_{PHASE}} {Z} \ = \frac {239.6V} {10Ω} = 24.0A$$ $$ P\ = I_{PHASE}^2\ R = (24.0A)^2\ \times \ 8Ω\ =\ 4.59kW$$ $$ P_T\ = 3\ P = 3\ \times \ 4.59kW\ =\ 13.8kW$$

Real power (and Reactive and Apparent Power) is the same for the Wye and Delta connected loads.


$$\theta = \tan \frac {6 \Omega}{8 \Omega} = 36.87°$$

Line quantities: $$\begin{align} P_T & = \sqrt {3}\ V_{LINE}\ I_{LINE}\ cos\ θ \\ & = \sqrt {3} \times 415V \times 24A \times cos\ 36.87° \\ & = 13.8kW \end{align}$$

Phase quantities: $$\begin{align} P_T & = 3 \ V_{PHASE}\ I_{PHASE}\ cos\ θ \\ & = 3 \times 239.6V \times 24A \times cos \ 36.87° \\ & = 13.8kW \end{align}$$


In a star connection I phase = I line. In a delta connection this is untrue because I line will be \$\sqrt3\$ higher due to two loads sharing each wire. This of course assumes a balanced supply and load.

Also, in your calculations you have said I line is 41.5 amps for the delta case - this is untrue - it is Iph that is 41.5 amps. However, ignoring that slight typo you have calculated both powers correctly it seems.


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