Question 118: Current Electricity

Question

A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5$$\Omega$$. When a resistance of 15$$\Omega$$ is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ___________$$\Omega$$.

Accepted Answer

Let the emf is E and internal resistance is r of this secondary cell so
$${{RE} \over {r + R}} \propto l$$
so $${{{R_1}E} \over {r + {R_1}}} \propto {l_1}$$
& $${{{R_2}E} \over {r + {R_2}}} \propto {l_2}$$
$$ \Rightarrow {{{R_1}(r + {R_2})} \over {{R_2}(r + {R_1})}} = {{{l_1}} \over {{l_2}}}$$
or $${{5(r + 15)} \over {15(r + 5)}} = {{200} \over {300}}$$
$$ \Rightarrow r = 5\,\Omega $$

Current Electricity

A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5$\Omega. When a resistance of 15\Omega is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ___________\Omega$.

Integer / Numeric Answer Type