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For a cell, the terminal potential difference is `2.2 V`, when circuit is open and reduces to `1.8 V `. When cell is connected to a resistance `R=5Omega`, the internal resistance of cell `(R)` isA. `(10)/(9) Omega`B. `(9)/(10) Omega`C. `(11)/(9) Omega`D. `(5)/(9) Omega` |
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Answer» Correct Answer - A In an open circuit, emd of cell `E = 2.2 V` In a closed circuit terminal potential difference `V =1.8 V` External resistance, `R = 5 Omega` Thus, internal resistance of cell is `r = ((E)/(V) -1) R = ((2.2)/(1.8)-1)5` `=((11)/(9)-1)5=(2)/(9)xx5 =(10)/(9)Omega`. |
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