Question

a resistance of 10 m long potentiometer wire is 10 ohm .if the current through it is 0.4A , what is the balancing length when two cells of emf 1.3 V & 1.1 V are connected so as to oppose each other​

Answer 1

Answer:

the balancing length is 50 cm when two cells of emf 1.3 V & 1.1 V are connected so as to oppose each other​

Explanation:

As we know that the two cells are opposing each other

So here we can say that the net potential difference across the two cells is given as

$V = V_1 - V_2$

so we will have

$V = 1.3 - 1.1$

$V = 0.2 Volts$

now total potential difference across the wire of potentiometer is given as

$V = iR$

$V = 0.4 (10) = 4 Volts$

total length of the wire is L = 10 m

now potential drop per unit length of the wire is given as

$\frac{V}{L} = 0.4 Volts/m$

now the balancing length is given as

$0.4 x = 0.2$

$x = 0.5 m$

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Topic : Potentiometer

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