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A cell has emf of 2.2 V, when connected to a resistance of `5 Omega`, the potential difference between the terminals of the cell becomes 2.1 V, the internal resistance for the cell isA. `0.12 Omega`B. `0.48 Omega`C. `0.24 Omega`D. `0.50 Omega` |
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Answer» Correct Answer - C `E=2.2 V, R=5 Omega, V=2.1 V`, `E=lR+ir implies (E-V)/(V)R=r` `(2.2-2.1)/(2.1)xx5=r implies (0.1)/(2.1)xx5=r` `r=0.24 Omega` |
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