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Standard XII

Physics

Conductivity and Resistivity

In the potent...

Question

In the potentiometer circuit suppose $ε$ =12V, $ℓ$ =1m
if the galvanometer current is zero when x=075m, (a) find the
potential difference V, (b) if v=emf of the battery, find its
internal resistance.

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Solution

Given $ε = 12 V$
$l = 1 m$
Potential Gradient $= k = \frac{ε}{l} = 12 \frac{V}{m}$
(a) In this situation $x = 0.75$ is null point
we know that
$ε = k l v = 12 \times 0.75 v = 12 \times \frac{3}{4} v = 9 V$

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In the potentiometer circuit suppose ε =12V, ℓ =1mif the galvanometer current is zero when x=075m, (a) find the potential difference V, (b) if v=emf of the battery, find its internal resistance.

Given ε=12V l=1m Potential Gradient =k=εl=12Vm(a) In this situation x=0.75 is null point we know thatε=klv=12×0.75v=12×34v=9V

In the potentiometer circuit suppose $ε$ =12V, $ℓ$ =1m
if the galvanometer current is zero when x=075m, (a) find the
potential difference V, (b) if v=emf of the battery, find its
internal resistance.

Given $ε = 12 V$
$l = 1 m$
Potential Gradient $= k = \frac{ε}{l} = 12 \frac{V}{m}$
(a) In this situation $x = 0.75$ is null point
we know that
$ε = k l v = 12 \times 0.75 v = 12 \times \frac{3}{4} v = 9 V$