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In the circuit shown the value of l in a...

In the circuit shown the value of l in ampere is

A

`1`

B

`0.60`

C

`0.4`

D

`1.5`

Text Solution

Verified by Experts

The correct Answer is:
C

So, net resistance,
`R = 2.4 + 1.6 = 4.0 Omega`
Therefore, current from the battery
`i = V/R = 4/4 = 1A`
Now, from the circuit (b)
`4I. = 6I`
`implies I. = 3/2 I`
But `i = I + I. - I + 3/2 I = 5/2 I`
`therefore I = 5/2 I`
`implies I = 2/5 = 0.4 A`
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Knowledge Check

  • In the circuit shown, the value of I in ampere is

    A
    1
    B
    `0.60`
    C
    0.4
    D
    1.5

  • In the circuit shown in figure value of V_(R) is

    A
    400V
    B
    200V
    C
    300V
    D
    zero

  • The value of R in the circuit shown in the figure is

    A
    `2Omega`
    B
    `3Omega`
    C
    `4Omega`
    D
    `1Omega`

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