Three resistors each of 2 Ω are connected together so that their total resistance is 3 Ω. Draw a diagram to show this arrangement and check it by calculation.

When two resistors (2 Ω each) in parallel combination are connected to the third resistor (2 Ω) in series connection then the resultant resistance is 3 Ω. The below diagram shows this arrangement of resistors:

To verify,

Total resistance of two resistors in parallel,

In parallel, if the equivalent resistance is R_P, then

$\dfrac{1}{R$P} = \dfrac{1}{R1} + \dfrac{1}{R2} \\[0.5em] \dfrac{1}{RP} = \dfrac{1}{2} + \dfrac{1}{2} \\[0.5em] \dfrac{1}{RP} = \dfrac{2}{2} \\[0.5em] \Rightarrow RP = 1 Ω

Hence, R_P = 1 Ω

In series connection,

Equivalent resistance = R_E = R_P + R₃
= 1 + 2 = 3 Ω

Hence, Equivalent resistance = 3 Ω