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Sagot :
Explanation:
10 /9 Ω
potential difference across the cell in open circuit is the emf of the cell.
Hence, emf E=2.2V
when, circuit is closed, potential difference across cell is given by V=E−Ir
And,
I= E/ R+r
Hence, V= E− Er/ R+r
⟹ V= ER/ R+r
⟹ 1.8= 2.2×5 /5+r
⟹9+1.8r=11
⟹ r= 2/ 1.8 Ω
⟹ r= 10/9 Ω
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