Question

How can three resistors of resistance 2ohm,3ohm and 6ohm be connected to go live total resistance of (a) 4ohm,(b)1ohm?​

Answer 1

$\sf\large \green{\underbrace{\red{Answer⋆}}}:$

(a) R2, R3 are parallel and series with R1

(b) R1, R2 and R3 are in parallel

Explanation:

$\sf R_1 = 2 \: ohm\\ \sf R_2 = 3 \: ohm \\ \sf R_3 = 6 \: ohm$

(a)

$\sf R_2 \: and \: R_3 \: are \: in \: parallel \\ \sf so \: let \: the \: total \: of \: R_2 \: and \: R_3 \: be \: R_a$

$\sf \large \frac{1}{R_a} = \frac{1}{R_2} + \frac{1}{R_3} \: as \: they \: are \: in \: parallel$

$\sf R_a = \frac{1}{3} + \frac{1}{6} \\ \\ \sf \frac{1}{R_a} = \frac{2 + 1}{6} \\ \\ \sf \frac{1}{R_a} = \frac{3}{6} \\ \\ \sf \frac{1}{R_a} = \frac{1}{2} \\ \\ \sf R_a = 2 \: ohm$

Ra and R1 is in series

And there total will be R

$\sf R = R_a + R_1 \\ \\ \sf R = 2 + 2 \\ \\ \sf R = 4 \: ohm$

(b)

$\sf R_1 R_2 \: and \:R_3 \: are \: in \: parallel \: so \: total \: be \: R$

$\sf \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \\ \\ \sf \frac{1}{R} = \frac{1}{2} + \frac{1}{3} + \frac{1}{ 6} \\ \\ \sf \frac{1}{R} = \frac{3+ 2 +1 }{6} \\ \\ \sf \frac{1}{R} = \frac{6}{6} \\ \\ \sf R = 1 \: ohm$