A uniform metallic wire having resistance 4 Ω is bent to form a square loop (ABCD) (see figure). A resistance of 2 Ω is connected between points B and D and a battery of 2 V is connected across points A and C as shown in the figure. Now the value of current (I) is : [figure]
- 4 A
- 4.5 A
- 8 A
- 2 A
Answer
(4) 2 A
Solution
Each side of the square has resistance 44 = 1 Ω.
Between A and C, the points B and D are symmetric (each is 1 Ω from A and 1 Ω from C), so they are at the same potential: no current in the 2 Ω (balanced Wheatstone bridge).
RAC = (1 + 1)(1 + 1)(1 + 1) + (1 + 1) = 1 Ω
I = 21 = 2 A
Answer: (4)