Question 577: Current Electricity

Question

To determine the internal resistance of a cell by using a potentiometer, the null point is at $$1 \mathrm{~m}$$ when shunted by $$3 \Omega$$ resistance and at a length $$1.5 \mathrm{~m}$$, when cell is shunted by $$6 \Omega$$ resistance The internal resistance of the cell is

Accepted Answer

$$\begin{aligned}
& \mathrm{r}=\mathrm{R}\left(\frac{\ell_1}{\ell_2}-1ight)=\mathrm{R}^{\prime}\left(\frac{\ell_1}{\ell_2^{\prime}}-1ight) \
& \ell_2=1 \mathrm{~m}, \ell_2^{\prime}=1.5 \mathrm{~m} \mathrm{R}=3 \Omega, \mathrm{R}^{\prime}=6 \Omega \
& 	herefore 3\left(\frac{\ell_1}{1}-1ight)=6\left(\frac{\ell_1}{1.5}-1ight)
\end{aligned}$$
Solving we get $$\ell_1=3 \mathrm{~m}$$
$$	herefore \quad r=3\left(\frac{3}{1}-1ight)=3 	imes 2=6 \Omega$$

Answer

$$1 \Omega$$

Answer

$$4 \Omega$$

Answer

$$2 \Omega$$

Answer

$$6 \Omega$$

Current Electricity

To determine the internal resistance of a cell by using a potentiometer, the null point is at $1 \mathrm{~m} when shunted by 3 \Omega resistance and at a length 1.5 \mathrm{~m}, when cell is shunted by 6 \Omega$ resistance The internal resistance of the cell is

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