Five equal resistances each of resistance R are connected as shown in the figure. A battery of V volts is connected between A and B. The current flowing in AFCEB will be [figure]
- 3VR
- VR
- V2R
- 2VR
Answer
(3) V2R
Solution
A is joined to F and B is joined to E, so the battery acts across F and E.
Between F and E there are two paths: FCE (R + R) and FDE (R + R), with R across C and D.
RFCRCE = RFDRDE = 1 balanced Wheatstone bridge, no current in CD
Req = (2R)(2R)2R + 2R = R I = VR
Current divides equally between FCE and FDE.
IAFCEB = I2 = V2R
Answer: (3)