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First a set of n equal resistors of R ea...

First a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistance R. A current I is observed to flow. Then the n resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is 'n'?

Text Solution

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When n resistors of value R are connected in series with cell with emf `epsilon` and internal resistance R then current .
`I = (E)/(R + nR) = (E)/(R (1 + n)) " "` .... (1)
Instead n resistor of R value are connected in parallel then current,
10 I = `(E)/(R + (R)/(n))`
`= (nE)/(R (n + 1)) " " ` ... (2)
`therefore 10 I = n ((E)/(R (n + 1) ) ) `
`therefore 10 I = nI " " therefore n = 10 `
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Knowledge Check

  • A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' ,and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors arc connected in parallel to the same battery. Then the current drqwn from battery becomes 10 I. The value of 'n' is

    A
    9
    B
    10
    C
    20
    D
    11

  • n resistors each of resistance r are connected to a battery of emf epsilon and internal resistance r. Then the ratio of terminal voltage to emf of battery = ...... .

    A
    n
    B
    `(n)/(n + 1)`
    C
    `(1)/(n+ 1)`
    D
    `(n+ 1)/(n)`

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