The equivalent resistance between the points | vedclass

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(D) Let the nodes be labeled. The $8\, \Omega$ resistor is in parallel with the $10\, \Omega$ resistor because they are connected between the same two

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let the nodes be labeled. The $8\, \Omega$ resistor is in parallel with the $10\, \Omega$ resistor because they are connected between the same two nodes. Equivalent resistance of $8\, \Omega$ and $10\, \Omega$ in parallel is $R_p = \frac{8 	imes 10}{8 + 10} = \frac{80}{18} = \frac{40}{9}\, \Omega$. Now,this $R_p$ is in series with the $6\, \Omega$ resistor. $R_s = \frac{40}{9} + 6 = \frac{40 + 54}{9} = \frac{94}{9}\, \Omega$. This $R_s$ is in parallel with the $4\, \Omega$ resistor. $R_{p2} = \frac{(\frac{94}{9}) 	imes 4}{(\frac{94}{9}) + 4} = \frac{\frac{376}{9}}{\frac{94 + 36}{9}} = \frac{376}{130} = \frac{188}{65}\, \Omega$. Finally,this is in series with the $7\, \Omega$ resistor. $R_{AB} = \frac{188}{65} + 7 = \frac{188 + 455}{65} = \frac{643}{65}\, \Omega$. Since this value is not among the options,the correct answer is $(D)$.

The equivalent resistance between the points $A$ and $B$ is:

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