Question

1. Two cells of emf E₁ and E₂ with internal resistances r₁ and r₂ respectively, are connected in parallel by connecting their positive terminals together and negative terminals together. Deduce an expression for equivalent emf and equivalent internal resistance of the combination.
2. A parallel combination, as stated in (a) above, of two cells of emfs Е and 3E and internal resistances R each is connected across a resistance 2R. Find the current that flows through resistance 2R.

Answer 1

a. Cells in parallel share the same terminal voltage.

Let,

Equivalent emf = E

Equivalent internal resistance = r

Using the current division principle:

E = `(E_1/r_1 + E_2/r_2)/(1/r_1 + 1/r_2)`

Multiply numerator and denominator by r₁r₂:

E = `(E_1 r_2 + E_2 r_1)/(r_1 + r_2)`

Parallel combination of internal resistances:

`1/r = 1/r_1 + 1/r_2`

r = `(r_1 r_2)/(r_1 + r_2)`

b. Using formula:

E_eq = `(E * R + 3E * R)/(R + R)`

= `(4 E R)/(2 R)`

= 2 E

Equivalent internal resistance (r) = `(R * R)/(R + R)`

= `R/2`

Total resistance (R_total) = `2 R + R/2`

= `(5 R)/2`

Current through 2R:

I = `E_"eq"/R_"total"`

= `(2 E)/(5 R//2)`

= `(4 E)/(5 R)`