Question

Equivalent resistance of the given network between points A and B is

• A. 31 /5$\sigma_{1} + \sigma_{2}$
• B. 41/5$\frac{\sigma_{1} + \sigma_{2}}{2}$
• C. 36/5/5$\sqrt{\sigma_{1}\sigma_{2}}$
• D. 49/5$\frac{2\sigma_{1}\sigma_{2}}{\sigma_{1} + \sigma_{2}}$

Answer 1

Answer: (C)

36/5/5$\sqrt{\sigma_{1}\sigma_{2}}$

Answer 2

: In each segment of the combination $3\Omega$and $2\Omega$resistances are connected in series separately.

$\therefore R' = 3 + 3 = 6\Omega$ and $R'' = 2 + 2 = 4\Omega$

$R'$ and $R''$ are connected in parallel

$\therefore$ For first segment

$\frac{1}{R_{eq1}} = \frac{1}{6} + \frac{1}{4} = \frac{2 + 3}{12} = \frac{5}{12}$

$R_{eql} = \frac{12}{5}\Omega$ similarly for second and third segment $R_{eq2}\frac{12}{5}\Omega$ and $R_{eq3} = \frac{12}{5}\Omega$

Now segment is connected in series then the total resistance of combination is.

$R_{eq} = R_{eq1} + R_{eq2} + R_{eq3}$

$\therefore R_{eq} = \frac{12}{5} + \frac{12}{5} + \frac{12}{5} = \frac{36}{5}\Omega$