A battery when connected to a 3 ohm load produces a current of 5 A. when the load is given a 5 ohm resistor in series, the current drops to 2.5 A. The resistance in the battery is
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R₁ = 3 Ω → i₁ = 5 A
R₂ = (3+5) = 8 Ω → i₂ = 2,5 A
r = ___ ?
i = E / (R + r)
E = i (R + r)
E = 5 · (3 + r)
E = 2,5 · (8 + r)
E = E
5 · (3 + r) = 2,5 · (8 + r)
15 + 5 r = 20 + 2,5 r
2,5 r = 5
r = 2 Ω
R₂ = (3+5) = 8 Ω → i₂ = 2,5 A
r = ___ ?
i = E / (R + r)
E = i (R + r)
E = 5 · (3 + r)
E = 2,5 · (8 + r)
E = E
5 · (3 + r) = 2,5 · (8 + r)
15 + 5 r = 20 + 2,5 r
2,5 r = 5
r = 2 Ω
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