When potential difference across the terminals of a battery is measured using an analogue voltmeter of resistance 95 ohms, the reading on the voltmeter is 5.70 V. When it is measured using a very high resistance digital meter the reading is 6.00 V. What is the internal resistance of the battery?

Answer 1

in the 1st case ... the internal voltage drop is 0.30 v
... the current is ... 5.70 v / 95 Ω

internal resistance ... 0.30 v / current

Answer 2

Thank you but I’m still a little confused.. what would the current be in this case?

Answer 3

Oh I think I understand, when I solve it I get 5 ohms as an answer, is that correct?