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The emfs of three cells connected in par...

The emfs of three cells connected in parallel are `E_(1) = 5 V, E_(2) = 8V and E_(3) =10 V and ` their internal resistances are `R_(1)=1 Omega , R_(2) = 2 Omega and R_(3) =3 Omega ,` respectively. By changing `E_(3)` to `E _(3N),` the equivalent emf is doubled, then `E_(3N)` in V is

A

12

B

34

C

47

D

82

Text Solution

Verified by Experts

The correct Answer is:
C
Equivalent emf and resistance in parallel combination is given by
`(E_(eq))/(r _(eq))= (E_(1))/( r_(1))+ (E_(2))/( r _(2)) + (E_(3))/(r_(3))`
`(E_(eq))/(r _(eq)) = 5/1 + 8/2 + 10/3 = 37/3"vole" " "...(i)`
Where `E_(3)` is replaced by `E_(3N),` equivalent emf gets doubled, so new equation
`(2E_(eq))/(r _(eq)) = 5/1+ 8/2+ (E_(3N))/(3) = (27 + E_(3N))/(3)`
`or (E_(eq))/(r _(eq)) = (27 + E _(3N))/(6) " " ...(ii)`
From Eqs (i) and (ii), we get
`implies (27 + E _(3N))/( 6 ) = (37)/(3)`
`implies 27+ E_(3N) = 74 implies E_(3N) = 47V`
`implies 27+ E_(3N) = 74 implies E_(3N) = 47V`
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Knowledge Check

  • The cells of emf's E_(1) and E_(2) be connector in parallel in a circuit. Let r_(1) and r_(2) be the internal resistance of the cells. Then the current through the circuit is

    A
    `(E_(1)+E_(2))/((r_(1)+r_(2)))`
    B
    `(E_(1)-E_(2))/((r_(1)+r_(2)))`
    C
    `(E_(1)+E_(2))/((r_(1)-r_(2)))`
    D
    `(2(E_(1)+E_(2)))/((r_(1)+r_(2)))`

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