Question

The equivalent resistance of the network shown in the figure
between the points A and B is
(A) 6Ω
(Β) 8Ω
(C) 16Ω
(D) 24 Ω​

Answer 1

Answer: Alternative B.

$R_{1} , R_{2}$ and $R_{3}$ in parallel:

$\frac{1}{Req} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3}\\\\\frac{1}{Req} = \frac{1}{8} + \frac{1}{16} + \frac{1}{16}\\\\\frac{1}{Req} = \frac{2 + 1 + 1}{16}\\\\\frac{1}{Req} = \frac{4}{16}\\\\Req = \frac{16}{4}\\\\Req_{1,2,3} = 4ohms$

$R_{4}$ and $R_{5}$ in parallel:

$Req = \frac{R4.R5}{R4+R5} \\\\Req = \frac{9.18}{9+18}\\\\Req = \frac{162}{27}\\\\Req_{4,5} = 6ohms$

Now we have $R_{1,2,3}$ in series with $R_{7}$:

$Req_{1,2,3,7} = R_{1,2,3} + R_{7}\\\\Req_{1,2,3,7} = 4 + 20\\\\Req_{1,2,3,7} = 24ohms$

And $Req_{4,5}$ in series with $R_{6}$:

$R_{4,5,6} = Req_{4,5} + R_{6} \\\\R_{4,5,6} = 6 + 6\\\\R_{4,5,6} = 12ohms$

Finally, we have $Req_{1,2,3,7}$ in parallel with $Req_{4,5,6}$:

$Req_{ab} = \frac{Req_{1,2,3,7}.Req_{4,5,6}}{Req_{1,2,3,7}+Req_{4,5,6} } \\\\Req_{ab} = \frac{12.24}{12+24}\\\\Req_{ab} = \frac{288}{36} \\\\Req_{ab} = 8ohms$