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For a circuit shown in fig find the valu...

For a circuit shown in fig find the value of resistance `R_(2)` and current `I_(2)` flowing thorugh `R_(2)`

Text Solution

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If equivalent resistance of parallel
combination of `R_(1)` and `R_(2)` is `R`, then
`R=(R_(1)R_(2))/(R_(1)+R_(2))=(10R_(2))/(10+R_(2))`
According to Ohm's law `R=(V)/(I)`
`R=(50)/(10)=5Omegaimplies(10R_(2))/(10+R_(2))=5impliesR_(2)=10Omega`
The current is equally divided into `R_(1)` and `R_(2)` Hence `I_(2)=5A`
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Knowledge Check

  • In the circuit shown in the given figure, the resistances R_(1) and R_(2) are respectively

    A
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    B
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    C
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  • In the circuit shown in the figure (A), R_(3) is a variable resistance As the value R_(3) is changed, current I though the cell varies as shown. Obvioulsy, the variation is asymptotic, i.e. Irarr6A as R_(3)rarroo . Resistance R_(1) and R_(2) are, respectively

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    `4Omega, 2Omega`
    B
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    C
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    D
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  • In the circuit shown in figure Current thorugh R_(1) is independent of

    A
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    B
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