Figure shows a potentiometer circuit for comparison of two resistances

1.	Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistance R = 10.0Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell ε.
Answer» In first case, resistance R is in parallel with cell ε, so p.d. across R = ε i.e., ε = RI ...(i) In second case, X is in parallel with cell so p.d. across X = ε i.e., ε = XI ...(ii) Let k be the potential gradient of potentiometer wire. If l1 and l2 are the balancing length corresponding to resistance respectively, then ε = kl₁ ...(iii) ε = kl₂...(iv) From (i) and (iii) RI = kl₁ ...(v) From (ii) and (iv) XI = kl₂ ...(vi) Dividing (vi) by (v), we get \(\frac{X}{R}=\frac{l_2}{l_1}\) \(\Rightarrow X=\frac{l_2}{l_1}R\) Here, R = 10.0 Ω, l1 = 58.3 cm, l2 = 68.5 cm ∴ \(X=\frac{68.5}{58.3}\times10.0=11.75\Omega\) If we fail to find the balance point with the given cell ε, then we shall take the driver battery (B1) of higher emf than given emf (ε).

Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistance R = 10.0Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell ε.

Answer»

In first case, resistance R is in parallel with cell ε, so p.d. across R = ε

i.e., ε = RI ...(i)

In second case, X is in parallel with cell so p.d. across X = ε

i.e., ε = XI ...(ii)

Let k be the potential gradient of potentiometer wire. If l1 and l2 are the balancing length corresponding to resistance respectively, then

ε = kl₁ ...(iii)

ε = kl₂...(iv)

From (i) and (iii) RI = kl₁ ...(v)

From (ii) and (iv) XI = kl₂ ...(vi)

Dividing (vi) by (v), we get

\(\frac{X}{R}=\frac{l_2}{l_1}\) \(\Rightarrow X=\frac{l_2}{l_1}R\)

Here, R = 10.0 Ω, l1 = 58.3 cm, l2 = 68.5 cm

∴ \(X=\frac{68.5}{58.3}\times10.0=11.75\Omega\)

If we fail to find the balance point with the given cell ε, then we shall take the driver battery (B1) of higher emf than given emf (ε).