1.

The resistance 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

Answer»

1.5 A
2.8 A
0.7 A
2.1 A

Solution :The resistance `5Omega,2Omega,6Omega and 15Omega` are connected in CYCLIC ORDER as SHOWN in the figure.

Resistance of the upper arm = `5Omega+2Omega=7Omega`
Resistance of the lower arm = `15Omega+6Omega=21Omega`
Current through the upper arm
`I_(1)=(2.8Axx21Omega)/(7Omega+21Omega)=2.1A`
Hence, the current through the `2Omega` resistor is 2.1 A.


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