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When two cells of negligible internal resistances are equal ...
physics waec-physics-2020 ssce-exam-2020 waec-2020
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When two cells of negligible internal resistances are equal e.m.f. denoted by E1 and E2 are connected in parallel, the combined e.m.f. E is given by

A. E = E1 + E2

B. 1/E = 1/E1 + 1/E2

C. E = E1 = E2

D. E = E1 / E2

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peter
Asked by peterrep:1.5K
University of Lagos Nigeria
29 August 2020

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dp
Felix,
The answer is C
2 likes, 0 dislikes





1 answer(s)

Felix
Felix answeredrep:12
Nigeria
23 October 2020
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For e.m.f connected in parallel, the value of E = E= E2 = E= En

The answer is C

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dp
jekwu16, Unn
C
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