Question

Figure, shows a network of seven resistors number 1 to 7, each equal to 1Ω connection to a 4V battery of negligible internal resistance The current I in the circuit is.... 

(A) 0.5A 

(B) 1.5A 

(C) 2.0A 

(D) 3.5A

Answer 1

The correct option (D) 3.5A

Explanation:

Using detta of star conversion for portion ABD & BCE.

using formula, R₁ = {(RbRc)/(Ra + Rb + Rc)}

Value of each resistance will be {(1 × 1)/(1 + 1 + 1)} = (1/3)Ω

hence circuit can be drawn as

further reduction,

hence I = [4/{(1/3) + (1/3) + {(2/3) || (5/3)}}] = [4/{(2/3) + (10/21)}]

= [4 / (24/21)]

∴ I = 4 × (21/24) = 3.5A