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Potentiometer: emf & internal resistance of batteries


A battery with internal resistance of 1, space, \Omega and an unknown emf drives a current across a 9, space, \Omega resistor.
When a potentiometer with a wire start text, A, B, end text is connected across the battery, the balancing length is, start text, A, E, end text, equals, 80, space, start text, c, m, end text.
What do we need in order to find the emf of the battery, E, in the fewest steps of calculation?
Note: Let's begin from scratch, and avoid using any derived results.
Choose 1 answer: