JEE Mains · Physics · STD 12 - 3. current electricity
A potentiometer \(PQ\) is set up to compare two resistances as shown in the figure. The ameter \(A\) in the circuit reads \(1.0\, A\) when two way key \(K_3\) is open. The balance point is at a length \(l_1\, cm\) from \(P\) when two way key \(K_3\) is plugged in between \(2\) and \(1\) , while the balance point is at a length \(l_2\, cm\) from \(P\) when key \(K_3\) is plugged in between \(3\) and \(1\) . The ratio of two resistances \(\frac{{{R_1}}}{{{R_2}}}\) is found to be
- A \(\frac{{{l_1}}}{{{l_1} + {l_2}}}\)
- B \(\frac{{{l_2}}}{{{l_2} - {l_1}}}\)
- C \(\frac{{{l_1}}}{{{l_1} - {l_2}}}\)
- D \(\frac{{{l_1}}}{{{l_2} - {l_1}}}\)
Answer & Solution
Correct Answer
(D) \(\frac{{{l_1}}}{{{l_2} - {l_1}}}\)
Step-by-step Solution
Detailed explanation
When key is at point \(( 1)\) \(\mathrm{V}_{1}=\mathrm{i} \mathrm{R}_{1}=\mathrm{x} l_{1}\) When key is at \(( 3)\) \({{\rm{v}}_2} = {\rm{i}}\left( {{{\rm{R}}_1} + {{\rm{R}}_2}} \right) = {\rm{x}}{l_2}\)…
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