Find the emf (V) and internal resistance...

Find the emf (`V`) and internal resistance `(R)` of a single battery which is equivalent toa parallel combination of two batteries of emf `V_1` and `V_2` and internal resistances `r_1` and `r_2` respectively, with polrities as shown in figure

Text Solution

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The correct Answer is:

A, B, C

a. Equivalent emf (`V`) of the battery
`PD` across the terminals of the battery is equal to its emf when current drawn from the battery is zero. In the given circuit,

Current in the circuit
`i=("Net emf")/("Total resistance")=(V_1+V_2)/(r_1+r_2)`
Therefore, potential difference between `A` and `B` would be
`V_A-V_B=V_1-ir_1`
`:. V_A-V_B=V_1-((V_1+V_2)/(r_1+r_2))r_1=(V_1r_1=V_2r_1)/(r_1+r_2)`
so the equivalent emf of the battery is
`V=(V_1r_2-V_2r_1)/(r_1+r_2)`
Note that if `V_1r_2=V_2r_1:V=0`If `V_1r_2gtV_2r_1:V_A-V_B`= Positive i.e. `A` side of the equivalent battery willl become the positive terminal and vice versa.
b. Internal resistance (r) of the battery
`r_1` and `r_2` are in parallel. Therefore , the internal resistance `r` will be given by
`1//r=1//r_1+1//r_2`
or `r=(r_1r_2)/(r_1+r_2)`

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the equivalent emf `epsi_(eq)` is smaller than `epsi_1`

the `epsi_(eq)` is given by `epsi_(eq) = epsi_1 + epsi_2` always .

`epsi_(eq)` is independent of internal resistances `r_1` and `r_2`

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.

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)`

the equivalent emf `epsi_(eq)` is smaller than `epsi_(1)`

The `epsi_(eq)` is given by `epsi_(eq)=epsi_(1)+epsi_(2)` always.

`epsi_(eq)` is independent of internal resistance `r_(1) and r_(2)`.

A resistor R is conneted to a parallel combination of two identical batteries each with emf E and an internal resistance r. The potential drop across the resistance R is.

`(2ER)/(2R+r)`

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`(2ER)/(R+2r)`