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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 `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

Text Solution

Verified by Experts

The correct Answer is:
D
`I =(21)/(21 + 7) xx 2.8`

I = 2.1 A
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Knowledge Check

  • 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

    A
    1.5 A
    B
    2.8 A
    C
    0.7 A
    D
    2.1 A

  • Current through the 5Omega resistor is

    A
    2A
    B
    4A
    C
    zero
    D
    1A

  • In the figure shown the current through 2 Omega resistor is

    A
    `2A`
    B
    `0A`
    C
    `4A`
    D
    `6A`