Six lead-acid type of secondary cells, each of emf 2.0 V and internal resistance 0.015 Ω Ω , are joined in series to provide a supply to a resistance of 8.5 Ω Ω . Determine:
(i) the current drawn from the supply and
(ii) its terminal voltage.
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
[(i) 
= 1.4 A, (ii) 
= 11.9 V]
Sol. 
(i) current i = 
= 1.4 A
(ii) terminal voltage V = iR = 
× 8.5 = 11.9 V
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