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theres a part b i think you neglected to answer ;)
Hi jamesswaggernaut, darn, you're right. The process to solve it for part b) is exactly the same as what's shown for part a). Since the internal resistances are in series with the resistor in the same branch as the battery, the only difference compared with a) is to add one more ohm to and to . I've made a note about this in the quick answer, and thanks for spotting it.
Best wishes,
Mr. Dychko
i did the same thing and still get a wrong answer