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+1 vote
1.1k views
in Current Electricity by (430 points)
edited by

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) \( 2 \Omega \)image

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1 Answer

+2 votes
by (33.2k points)

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)

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