Question

Figure, shows a network of eight resistors numbered 1 To 8, each equal to 2Ω, connected to a 3V battery of negligible internal resistance The current I in the circuit is …. 

(A) 0.25A 

(B) 0.5A 

(C) 0.75A 

(D) 1.0A

Answer 1

The correct option (D) 1.0A 

Explanation:

Using delta star conversion for ABC & CDF,

hence each resistance will be (2/3)Ω

Hence (2/3) + 2 + (2/3) = (4/3) + 2 = (10 / 3)Ω

also (2/3) + 2 + (2/3) = (10 / 3)Ω

hence

∴ Equivalent R = (2/3) + (2/3) + [(10/3) || (10/3)]

= (4/3) + (5/3) = 3Ω

 ∴ I = (3/R) = (3/3) = 1A