Find the emf (V) and internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs V1 and V2 and internal resistances r1 and r2 respectively, with polarities as shown in figure

Question

Find the emf (V) and internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs V1 and V2 and internal resistances r1 and r2 respectively, with polarities as shown in figure (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(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 internal circuit,

i = \frac{N e t e m f}{T o t a l res i s t an ce} = \frac{V _{1} + V _{2}}{r _{1} + r _{2}}

Therefore, potential difference between A and B would be

V_{A} - V_{B} = V_{1} - i r_{1}

∴ V_{A} - V_{B} = V_{1} - (\frac{V _{1} + V _{2}}{r _{1} + r _{2}}) r_{1} = \frac{V _{1} r _{2} - V _{2} r _{1}}{r _{1} + r _{2}}

So, the equivalent emf of the battery is

V = \frac{V _{1} r _{2} - V _{2} r _{1}}{r _{1} + r _{2}}

Note that if

V_{1} r_{2} = V_{2} r_{1} : V = 0

If

V_{1} r_{2} > V_{2} r_{1} : V_{A} - V_{B} = P os i t i v e

i.e. A side of the equivalent battery will 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/ k + 1/ r_{2}

or

r = \frac{r _{1} r _{2}}{r _{1} + r _{2}}

Q. Find the emf (V) and internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs V1​ and V2​ and internal resistances r1​ and r2​ respectively, with polarities as shown in figure

Q. Find the emf (V) and internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs $V_{1}$ and $V_{2}$ and internal resistances $r_{1}$ and $r_{2}$ respectively, with polarities as shown in figure