Write the formulas for delta-to-star and star-to-delta transformation. Using them, convert a delta network with \(R_{AB}=3\ \Omega\), \(R_{BC}=6\ \Omega\) and \(R_{CA}=9\ \Omega\) into an equivalent star.

Answer: \(R_A=1.5\ \Omega,\ R_B=1\ \Omega,\ R_C=3\ \Omega\) (star arms from the centre N to A, B, C).

Delta to star: each star arm = (product of the two delta resistors meeting at that terminal) / (sum of all three):

\[R_A=\frac{R_{AB}R_{CA}}{R_{AB}+R_{BC}+R_{CA}},\quad R_B=\frac{R_{AB}R_{BC}}{R_{AB}+R_{BC}+R_{CA}},\quad R_C=\frac{R_{BC}R_{CA}}{R_{AB}+R_{BC}+R_{CA}}\]

Star to delta: each delta resistor = sum of the two star arms it joins + their product / the third arm:

\[R_{AB}=R_A+R_B+\frac{R_AR_B}{R_C},\quad R_{BC}=R_B+R_C+\frac{R_BR_C}{R_A},\quad R_{CA}=R_C+R_A+\frac{R_CR_A}{R_B}\]

Conversion: sum \(=3+6+9=18\ \Omega\).

\(R_A=\dfrac{3\times9}{18}=1.5\ \Omega\), \(R_B=\dfrac{3\times6}{18}=1\ \Omega\), \(R_C=\dfrac{6\times9}{18}=3\ \Omega\).

Check (back to delta): \(R_{AB}=1.5+1+\dfrac{1.5\times1}{3}=3\ \Omega\) ✓.