The emf of a Daniel cell is 1.08 V When the terminals of the cells are connected to a resistance of 3 Ω , the potential difference across the terminals is found to be 0.6 V . Then the internal resistance of the cell is

The emf of a Daniel cell is $1.08\text{\hspace{0.17em}}\mathrm{V}$ When the terminals of the cells are connected to a resistance of $3\text{\hspace{0.17em}}\mathrm{\Omega }$ , the potential difference across the terminals is found to be $0.6\text{\hspace{0.17em}}\mathrm{V}$ . Then the internal resistance of the cell is

1. A

$1.8\mathrm{\Omega }$

2. B

$2.4\mathrm{\Omega }$

3. C

$3.24\mathrm{\Omega }$

4. D

$0.2\mathrm{\Omega }$

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Solution:

emf of a Daniel cell  $E=1.08\mathrm{V}$ Resistance  $R=3\mathrm{\Omega }$
Potential difference $V=0.6\mathrm{V}$  Internal resistance  $\mathrm{r}$

$\mathrm{V}=\mathrm{E}-\mathrm{Ir}$

$\left(\frac{1.08}{\mathrm{R}+\mathrm{r}}\right)\mathrm{r}=0.48⇒\mathrm{r}=2.4\mathrm{\Omega }$

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