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The resistances in the four arms of a Wh...

The resistances in the four arms of a Wheatstone network in cyclic order are `5Omega, 2Omega, 6Omega and 15Omega`. If a current of 2.8 A enters the junction of `5Omega and 15Omega`, then the current through `2Omega` resistor is

A

1.5 A

B

2.8 A

C

0.7 A

D

2.1 A

Text Solution

Verified by Experts

The correct Answer is:
D
Current through
`2Omega=2.8{((15+6))/((5+2)+(15+6))}`
`=2.1A`
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