A) \[\frac{10}{9}\Omega \]
B) \[\frac{9}{10}\,\Omega \]
C) \[\frac{11}{9}\Omega \]
D) \[\frac{5}{9}\Omega \]
Correct Answer: A
Solution :
In an open circuit, emf 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 \[r=\left( \frac{E}{V}-1 \right)\,\,R=\left( \frac{2.2}{1.8}-1 \right)\,5\] \[=\left( \frac{11}{9}-1 \right)\,5=\frac{2}{9}\times 5=\frac{10}{9}\Omega \]You need to login to perform this action.
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