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find the emf (epsi(0)) and internal resi...


find the emf `(epsi_(0))` and internal resistance `(r_(0))` if a battery which is equivalent to a parallel combination of two batteries of emfs `epsi_(1) and epsi_(2)` and internal resistances `r_(1) and r_(2)` respectively, with polarities as shown in figure.

Text Solution

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The correct Answer is:
`epsi_(0)=(epsi_(1)r_(2)-epsi_(2)r_(1))/(r_(1)+r_(2)),r_(0)=(r_(1)r_(2))/(r_(1)+r_(2))`
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Knowledge Check

  • Two batteries of emf epsi_(1) and epsi_(2)(epsi_(2) gt epsi_(1)) and internal resistance r_(1) and r_(2) respectively are connected in parallel as shown in figure.

    A
    The equivalent emf `epsi_(eq)` of the two cells is between `epsi_(1) and epsi_(2)`, i.e., `epsi_(1) lt epsi_(eq) lt epsi_(2)`
    B
    the equivalent emf `epsi_(eq)` is smaller than `epsi_(1)`
    C
    The `epsi_(eq)` is given by `epsi_(eq)=epsi_(1)+epsi_(2)` always.
    D
    `epsi_(eq)` is independent of internal resistance `r_(1) and r_(2)`.
  • Two batteries of emf epsi_1 and epsi_2 (epsi_2 gt epsi_1) and internal resistances r__1 and r_2 respectively are connected in parallel as shown in Fig. Then

    A
    the equivalent emf `epsi_(eq)` of the two cells is between`epsi_1` and `epsi_2` i.e., `epsi_1 lt epsi_(eq) lt epsi_2`
    B
    the equivalent emf `epsi_(eq)` is smaller than `epsi_1`
    C
    the `epsi_(eq)` is given by `epsi_(eq) = epsi_1 + epsi_2` always .
    D
    `epsi_(eq)` is independent of internal resistances `r_1` and `r_2`
  • Two batteris of emf epsilon_(1) and epsilon_(2) wit respective internal resistance r_(1) and r_(2) are connected in parallel. Now

    A
    The effective emf of this parallel combination is `((epsi_(1)r_(1)+epsi_(2)r_(2)))/((r_(1)+r_(2)))`
    B
    The effective internal resistance of this combination is `(2r_(1)r_(2))/((r_(1)+r_(2)))`
    C
    the effective emf of this parallel combination is `((epsi_(1)r_(2)+epsi_(2)r_(1)))/((r_(1)+r_(2)))`
    D
    The effective internal resistance of this combination is `(r_(1)r_(2))/((r_(1)+r_(2)))`
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