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  • What is total resistance across AB in the following network. <div cl

What is total resistance across AB in the following network.

Option: 1

6.4 \ \Omega


Option: 2

2.4 \ \Omega


Option: 3

7.4 \ \Omega


Option: 4

5.4 \ \Omega


Answers (1)

best_answer

We can't apply Wheatstone's bridge condition because
\frac{4}{8} \neq \frac{2}{8}
\therefore Break delta \triangle \mathrm{ACD} into star  connection, For that assign 1,2 and 3 like following fig.

\begin{aligned} \mathrm{R}_1 & =\frac{\mathrm{R}_{12} \times \mathrm{R}_{13}}{\mathrm{R}_{12}+\mathrm{R}_{13}+\mathrm{R}_{23}} \\ & =\frac{4 \times 2}{4+2+4}=\frac{8}{10}=0.8 \Omega \\ \mathrm{R}_2 & =\frac{R_{12} \times R_{23}}{R_{12}+R_{23}+R_{13}}=\frac{4 \times 4}{4+2+4}=\frac{16}{10}=1.6 \Omega \end{aligned}\mathrm{R}_3=\frac{R_{23} \times R_{13}}{R_{12}+R_{23}+R_{13}}=\frac{4 \times 2}{4+2+4}=\frac{8}{10}=0.8 \Omega
Connect R_1, R_2 and R_3 like following fig.

remove delta circuit from above fig., we get


resistance across A and B is
\begin{aligned} R & =0.8+[(8+1.6) \|(0.8+8)] \\ & =0.8+\left[\frac{(9.6) \times 8.8}{(9.6+8.8)}\right]=5.4 \Omega \end{aligned}
 

 

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Gaurav

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