A) \[\frac{10}{9}\Omega \]
B) \[\frac{9}{10}\,\Omega \]
C) \[\frac{11}{9}\Omega \]
D) \[\frac{5}{9}\Omega \]
Correct Answer: A
Solution :
| [a] In an open circuit, emf of cell |
| \[E=2.2\text{ }V\] |
| In a closed circuit, terminal potential difference |
| \[V=1.8\text{ }V\] |
| External resistance, \[R=5\,\Omega \] |
| Thus, internal resistance of cell is |
| \[r=\left( \frac{E}{V}-1 \right)\,\,R=\left( \frac{2.2}{1.8}-1 \right)\,5\] |
| \[=\left( \frac{11}{9}-1 \right)\,5=\frac{2}{9}\times 5=\frac{10}{9}\Omega \] |
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