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A battery of emf $10 \mathrm{~V}$ and internal resistance $3 \Omega$ is connected to a resistor. If the current in the circuit is $0.5 \mathrm{~A}$, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
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Given, $\mathrm{E}=10 \mathrm{~V}, \mathrm{r}=3 \Omega, \mathrm{I}=0.5 \mathrm{~A}, \mathrm{R}=?, \mathrm{~V}=$ ?
By formula, $I=\frac{E}{R+r}$ and
$$
\mathrm{R}=(\mathrm{E} / \mathrm{I})-\mathrm{r}=(10 / 0.5)-3=20-3=17 \Omega
$$
By relation, $\mathrm{V}=\mathrm{IR}=0.5 \times 17=8.5 \mathrm{~V}$
By formula, $I=\frac{E}{R+r}$ and
$$
\mathrm{R}=(\mathrm{E} / \mathrm{I})-\mathrm{r}=(10 / 0.5)-3=20-3=17 \Omega
$$
By relation, $\mathrm{V}=\mathrm{IR}=0.5 \times 17=8.5 \mathrm{~V}$
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