Question

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

Answer 1

✔Answer:

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

$V = iR \\ V = 0.4 \: (10) = 4volts$

total length of the wire is L = 10 m

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

$\frac{V}{L} = 0.4 \: volt \: {m}^{ - 1}$

now the balancing length is given as

$0.4 x = 0.2 \\ \\ x = 0.5m.$

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