Question

Show how would you connect four resistors, each of resistance 1 Ω, so that the combination has a resistance of (i) 1 Ω (ii) 2.5 Ω

Answer 1

Hey mate..
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1) 1Ω

▪If 1 Ω and 1 Ω resistances of a two resistors are connected in Parallel and the equivalent resistance is taken as Rρ then,

=> 1 / Rρ = 1 / 1 + 1 / 1

=> 1 / Rρ = 2 / 1

=> Rρ = 1 / 2 Ω

Similarly we get,

Rρ' = 1 / 2 Ω ( °•° Other two resistors )

Now,

If Rρ αηd Rρ' are connected in series , then the Equivalent Resistance ( R ) will be :-

=> R = Rρ + Rρ '

=> R = 1/2 + 1/2

=> R = 2/2

=> R = 1 Ω [ Shown (i) ]

2) 2.5 Ω

▪ If 1 Ω and 1 Ω resistances of a two resistors are connected in Parallel and if the equivalent resistance is taken as Rς then,

=> 1 / Rς = 1 / 1 + 1 / 1

=> 1 / Rς = 2 / 1

=> Rς = 1 / 2 Ω

And,

▪If the other two resistors each of resistance 1 Ω are connected in series and if the equivalent resistance is taken as Rω then,

=> Rω = ( 1 + 1 ) Ω

=> Rω = 2Ω

Now, If Rς and Rω are connected in series and the equivalent resistance is taken as Rs we get,

=> Rs = ( 1 / 2 + 2 ) Ω

=> Rs = 5 / 2 Ω

=> Rs = 2.5 Ω

Hope it helps !!☺

Answer 2

Hola Friend ✋✋✋

(1) We should connect like in figure in the attachment...

$\frac{1}{ r_{12}} = \frac{1}{ r_{1} } + \frac{1}{ r_{2}}$

$r_{12} = \frac{1}{2}$

$\frac{1}{ r_{34}} = \frac{1}{ r_{3} } + \frac{1}{ r_{4}}$
$r_{34} = \frac{1}{2}$

$r_{1234} = \frac{1}{2} + \frac{1}{2} = 1ohm$


(2) 2.5 ohm

$\frac{1}{ r_{23}} = \frac{1}{ r_{2} } + \frac{1}{ r_{3}}$

$r_{23} = \frac{1}{2} = 0.5ohm$


$r_{1234} = r_{1} + r_{23} + r_{4}$


$r_{1234} = 2.5ohm$


Hope it helps