A potentiometer wire carries a steady current. The potential differenc

1.	A potentiometer wire carries a steady current. The potential difference across \( 70 cm \) length of it balances the potential difference across a \( 2 \Omega \) coil connected to a cell of emf \( 2.0 V \) and an unknown internal resistance \( r \). When a \( 1 \Omega \) coil is placed in parallel with the \( 2 \Omega \) coil, a length equal to \( 50 cm \) of the potentiometer wire is required to balance the potential difference across the parallel combination. What is the value of the internal resistance (r) of the cell? a) \( 0.5 \Omega \) b) \( 0.7 \Omega \) c) \( 0.3 \Omega \) d) \( 2 \Omega \)
Answer» Given l₁ = 70 cm l₂ = 50 cm E = 2 r = ? For balancing length l₁ = \((\frac{E}{R+r})\frac{R_p}{L}\times l_1\) \((\frac2{2+r})\frac{R_p}{L}\times70\).......(i) For balancing length l₂ _{\((\frac{E}{R+r})\frac{R_p}{L}\times l_1\)} _{\(\left(\cfrac{2}{\frac23+r}\right)\frac{R_p}{L}\times50\).......(ii)} Comparing equation (i) and (ii) \((\frac2{2+r})\frac{R_p}L\times70=\left(\cfrac{2}{\frac23+r}\right)\frac{R_p}L\times50\) \(\frac{140}{2+r}=\cfrac{100}{\frac{(2+3i)}{3}}\) \(\frac{140}{2+r}=\frac{300}{2+3r}\) 280 + 420 r = 600 + 300r 120r = 600 -280 120 r = 320 r = \(\frac{320}{120}\) r = 2.6 Ω (d)

A potentiometer wire carries a steady current. The potential difference across \( 70 cm \) length of it balances the potential difference across a \( 2 \Omega \) coil connected to a cell of emf \( 2.0 V \) and an unknown internal resistance \( r \). When a \( 1 \Omega \) coil is placed in parallel with the \( 2 \Omega \) coil, a length equal to \( 50 cm \) of the potentiometer wire is required to balance the potential difference across the parallel combination. What is the value of the internal resistance (r) of the cell? a) \( 0.5 \Omega \) b) \( 0.7 \Omega \) c) \( 0.3 \Omega \) d) \( 2 \Omega \)

Answer»

Given l₁ = 70 cm

l₂ = 50 cm

E = 2

r = ?

For balancing length l₁ = \((\frac{E}{R+r})\frac{R_p}{L}\times l_1\)

\((\frac2{2+r})\frac{R_p}{L}\times70\).......(i)

For balancing length l₂

_{\((\frac{E}{R+r})\frac{R_p}{L}\times l_1\)}

_{\(\left(\cfrac{2}{\frac23+r}\right)\frac{R_p}{L}\times50\).......(ii)}

Comparing equation (i) and (ii)

\((\frac2{2+r})\frac{R_p}L\times70=\left(\cfrac{2}{\frac23+r}\right)\frac{R_p}L\times50\)

\(\frac{140}{2+r}=\cfrac{100}{\frac{(2+3i)}{3}}\)

\(\frac{140}{2+r}=\frac{300}{2+3r}\)

280 + 420 r = 600 + 300r

120r = 600 -280

120 r = 320

r = \(\frac{320}{120}\)

r = 2.6 Ω (d)