Question

How can three resistors of resistances 2Ω, 3Ω and 6Ω be connected to
give a total resistance of 1Ω ?

Answer 1

Here, R1=2Ω,R2 =3Ω,R3 = 6Ω, and R =1Ω
Since the equivalent resistance of the combination is of lesser value than any of the resistors of the combination, it is clear that the resistors should be connected in parallel. It can be further confirmed by using the formula





Therefore, resistors should be connected in parallel.

Answer 2

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If we connect the given 3 resistors in the following way (Refer this attachment).


$\bold{Then,}$


We get net resistance as $1Ω.$


Here all the resistors are connected in $\bold{parallel.}$


$\bold{Therefore,}$


Their equivalent resistance will be given as:-


$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$


$\frac{1}{R} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$


$\frac{1}{R} = \frac{3 + 2 + 1}{6}$


$\frac{1}{R} = \frac{6}{6}$


$R = 1Ω$


Therefore, the net equivalent resistance of the circuit is $1Ω.$


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