A battery of emf 2 Vand internal resistance 0.5ohm is connected across to a resistance of 9.5ohm How many electron pass through a cross section of resistance in 1 second
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Answer:
EMF of the cell(E) =2 V
internal resistance(r)= 0.1 V
External resistance(R)=3.9 V
V=IR
=(E÷(R+r))×R
=(2÷(3.9+0.1))×3.9
=1.95 V
Therefore potential difference across the terminals of the cell =
1.95V.
Explanation:
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