Discussion
Two cells of emfs 2 V and 6V each with internal resistance of 1 Ω are connected in series to an external resistance 8 Ω . Then the current in the circuit is
Answer: D
The current is 0.8 A.
Connected in series (aiding), the emfs add and so do the internal resistances:
\[ I = \frac{2 + 6}{8 + 1 + 1} = \frac{8}{10} = 0.8\ \text{A} \]
Ignoring the internal resistances gives 1 A. If the cells opposed each other the net emf would be 4 V and the current 0.4 A, which is not offered.
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