When two cells of negligible internal resistances and equal e.m.f. denoted by E₁ and E₂ are connected in parallel, combined e.m.f. E is given by

Correct Answer is : E=E1=E2E=E₁=E₂E=E1=E2

Correct Answer is : E=E1=E2E=E₁=E₂E=E1=E2

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WAEC Physics 2020 Paper

Question : 49

Total: 51

E_1

E_2

E

Solution:

Concept:

When cells with equal e.m.f. and negligible internal resistance are connected in parallel, the combined e.m.f. is equal to the e.m.f. of each cell.

Explanation:

For a parallel connection, the positive terminals are joined together and the negative terminals are joined together.

So the potential difference across the combination is the same as the potential difference across each cell.

Since both cells have equal e.m.f.

E_1

and

E_2

, the effective e.m.f. is:

E = E_1 = E_2

Internal resistances are negligible, so no additional voltage drop or division occurs.

Therefore, the combined e.m.f. is not the sum or the reciprocal sum of the individual e.m.f.s.

Answer:

E = E_1 = E_2

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