Question 2.47: A delta-connected three-phase motor load is supplied from a ...

Line voltage, V_{L} = 400  V,  V_{Ph} = V_{L} for delta connection
Line current,                    I_{L} = \sqrt{3}  I_{Ph} = 21  A
Impedance of each winding,            Z_{Ph}=\frac{ V_{Ph}}{ I_{Ph}}=\frac{400}{I_{L}/ \sqrt{3}}= \frac{400  ×  \sqrt{3}}{21 }=33  Ω

P = \sqrt{3}  V_{L}  I_{L}  \cosΦ= \sqrt{3}  × 400 × 21 × \cosΦ

or,     \cosΦ = \frac{P}{\sqrt{3}  ×  400  ×  21 }= \frac{ 11  ×  1000}{\sqrt{3}  ×  400  ×  21}=0.756
When the motor windings are star connected

V_{Ph}=\frac{V_{L}}{\sqrt{3}}=\frac{ 400}{1.732}=231  V

Z_{Ph} will remain the same as the same windings are connected in star

I_{Ph}=\frac{V_{Ph}}{Z_{Ph}}=\frac{ 231}{33}=7  A

In star connection, the line current is the same as phase current, hence, I_{L}=I_{Ph}= 7  A.
Power factor, depends on the circuit parameters

Since both R and Z remain unchanged, the power factor will remain the same at 0.756. Students may note that line current in a star connection is one-third of the line current in a delta connection.
or,    R_{1} \ ^{2} = 138
or,     R_{1} = 11.74  Ω