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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 - Physics

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ConceptPotentiometer

Question

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) ε12 and (ii) position of null point for the cell ε1 only.

Solution

(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:

\[\epsilon_1 - \epsilon_2 = \left( 120x \right)I . . . (1)\] and
\[\epsilon_1 + \epsilon_2 = \left( 300x \right)I . . . (2)\]
Dividing both equations, we get:
 

\[\frac{\epsilon_1 - \epsilon_2}{\epsilon_1 + \epsilon_2} = \frac{120}{300}\]

\[ \Rightarrow 180 \epsilon_1 = 420 \epsilon_2 \]

\[ \Rightarrow \frac{\epsilon_1}{\epsilon_2} = \frac{7}{3}\]

(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

  Is there an error in this question or solution?

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Solution 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 Concept: Potentiometer.
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