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

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

A

1.5 A

B

2.8 A

C

0.7 A

D

2.1 A

Verified by Experts

The correct Answer is:

D

`I =(21)/(21 + 7) xx 2.8`

I = 2.1 A

Similar Questions

Explore conceptually related problems

Find the current through 12 Omega resistor in

Current through the 5Omega resistor is

The value of current through 2 Omega resistor is

In the figure shown the current through 2 Omega resistor is

Electric current through 400 Omega resistor is :

In the arrangement the current through 5 Omega resistor is

Current flowing through 4 Omega resistor is :

Find the current through 2 Omega and 4 Omega resistance.

Four resistances arranged to form a Wheatstone network are 8Omega, 12Omega,6Omega and 27Omega resistance, so that the bridge is balanced, is