# Physics

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?

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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

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

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

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