The star connection by the resistors \(3\ \mathrm{\Omega}\),\(4\ \mathrm{\Omega}\) and\(6\ \mathrm{\Omega}\) is converted to delta connection, we have by the redraw of Fig. 14

\(R_1=\frac{6\times3+3\times4+4\times6}{3}=18\ \mathrm{\Omega}\)

\(R_2=\frac{6\times3+3\times4+4\times6}{4}=13.5\ \mathrm{\Omega}\)

\(R_3=\frac{6\times3+3\times4+4\times6}{6}=9\ \mathrm{\Omega}\)

Fig. 14(b)

Fig. 14(c)

\(R_4=\frac{9\times18}{9+18}=6\ \mathrm{\Omega}\)

\(R_4=\frac{1\times13.5}{1+13.5}=0.93\ \mathrm{\Omega}\)

\(R_4=\frac{9\times1}{9+1}=0.9\ \mathrm{\Omega}\)

Hence the resistance between terminals A and B is \(6\ \mathrm{\Omega}\)