The current passing through the battery in the given circuit, is: [figure]
- 2.5 A
- 1.5 A
- 2.0 A
- 0.5 A
Answer
(4) 0.5 A
Solution
The wire AF joins A and F, so between B and E the network is a Wheatstone bridge: AB = 5 , AE = 3 , BD = 2.5 (through C), DE = 1.5 and the 6 resistor joins A and D.
Check the balance condition:
RABRAE = 53 and RBDRDE = 2.51.5 = 53
The bridge is balanced, so no current flows through the 6 resistor.
RBE = (5+3)(2.5+1.5)(5+3)+(2.5+1.5) = 8 412 = 83
This is in series with 1.5 , 5.5 and 13 :
Rtotal = 83 + 1.5 + 5.5 + 13 = 10
I = ERtotal = 510 = 0.5 A
Answer: (4)