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Question:
Published on: 21 June, 2022

A network of resistances is formed as shown in Fig.14. Compute the resistance between the points A and B.

Fig. 14(a)

Answer:

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 \(6\ \mathrm{\Omega}\)

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