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प्रश्न
Figure shows a long potentiometer wire AB having a constant potential gradient. The null points for the two primary cells of emfs ε1 and ε2 connected in the manner shown are obtained at a distance of l1 = 120 cm and l2 = 300 cm from the end A. Determine (i) ε1/ε2 and (ii) position of null point for the cell ε1 only.
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उत्तर
(i) Let x be the resistance per unit length of the potentiometer wire and I be the constant current flowing through it. Then from the figure, we have:
\[\frac{\epsilon_1 - \epsilon_2}{\epsilon_1 + \epsilon_2} = \frac{120}{300}\]
\[ \Rightarrow 180 \epsilon_1 = 420 \epsilon_2 \]
\[ \Rightarrow \frac{\epsilon_1}{\epsilon_2} = \frac{7}{3}\]
(ii) Now adding equations (1) and (2), we get:
\[2 \epsilon_1 = \left( 420x \right)I\]
\[ \Rightarrow \epsilon_1 = \left( 210x \right)I\]
Comparing with
\[\epsilon = \left( Lx \right)I\]we get:
length of balancing point, L = 210 cm for cell 1
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