In a long uniform potentiometer wire AB of constant potential gradient, the null points for two primary cells of EMFs $\varepsilon_1$ and $\varepsilon_2$ (connected as shown) are obtained at $120\,\text{cm}$ and $300\,\text{cm}$ from A. Find (i) $\varepsilon_1/\varepsilon_2$ and (ii) the position of the null point for $\varepsilon_1$ alone. How is the sensitivity of a potentiometer increased?

(i) $\dfrac{\varepsilon_1}{\varepsilon_2}=\dfrac{7}{3}$. (ii) It follows from $\varepsilon_1=k\ell$; sensitivity is increased by reducing the potential gradient (longer wire / smaller driving current).

CBSE Class 12 · 2012 · Current ElectricityShort Answer

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In a long uniform potentiometer wire AB of constant potential gradient, the null points for two primary cells of EMFs

ε_{1}

and

ε_{2}

(connected as shown) are obtained at

120 cm

and

300 cm

from A. Find (i)

ε_{1} / ε_{2}

and (ii) the position of the null point for

ε_{1}

alone. How is the sensitivity of a potentiometer increased?

Solution

In one connection the cells oppose,

ε_{1} - ε_{2} = k (120)

; in the other they add,

ε_{1} + ε_{2} = k (300)

. Dividing,

\frac{ε _{1} - ε _{2}}{ε _{1} + ε _{2}} = \frac{120}{300} = \frac{2}{5}

, which gives

5 (ε_{1} - ε_{2}) = 2 (ε_{1} + ε_{2}) \Rightarrow 3 ε_{1} = 7 ε_{2}

, so

\frac{ε _{1}}{ε _{2}} = \frac{7}{3}

. Sensitivity rises when the potential gradient

k

is made small.

CBSE Class 12 Board · 2012 · Subjective

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CBSE Class 12 · 2012 · Current ElectricityShort Answer

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