For the network shown in the figure the value of the current i is [figure]
- 9V35
- 18V5
- 5V9
- 5V18
Answer
(4) 5V18
Solution
Arms: 4 and 2 in the upper path, 6 and 3 in the lower path, 4 as the bridge.
42 = 63 balanced Wheatstone bridge, no current in the middle 4
Req = (4 + 2)(6 + 3)(4 + 2) + (6 + 3) = 5415 = 185
i = VReq = 5V18
Answer: (4)