Two resistor of 2Ω in series parallel to resistor 5Ω in series to a 2Ω resistor. This configuration gives to us an equivalent resistor of 2.55Ω.
To solve this problem we have to use the rules of conection of resistor in series and parallel.
A resistor R1 in serie with other resistor R2 gives us an equivalent resistor Req= R1 + R2.
A resistor R1 in parallel with other resistor R2 gives us an equivalent resistor Req = R1.R2/R1+R2.
The circuit that show an arregement of resistor which we obtain a equivalent resistor of 2.5Ω from three resistor of 2Ω and 5Ω respectively is attached in the image:
