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 ___________$\Omega$.
Answer (integer)
5
Solution
<p>Let the emf is E and internal resistance is r of this secondary cell so</p>
<p>${{RE} \over {r + R}} \propto l$</p>
<p>so ${{{R_1}E} \over {r + {R_1}}} \propto {l_1}$</p>
<p>& ${{{R_2}E} \over {r + {R_2}}} \propto {l_2}$</p>
<p>$$ \Rightarrow {{{R_1}(r + {R_2})} \over {{R_2}(r + {R_1})}} = {{{l_1}} \over {{l_2}}}$$</p>
<p>or ${{5(r + 15)} \over {15(r + 5)}} = {{200} \over {300}}$</p>
<p>$\Rightarrow r = 5\,\Omega$</p>
About this question
Subject: Physics · Chapter: Current Electricity · Topic: EMF and Internal Resistance
This question is part of PrepWiser's free JEE Main question bank. 120 more solved questions on Current Electricity are available — start with the harder ones if your accuracy is >70%.