A potentiometer wire of length 1 m has a resistance of 10$\Omega$. It is connected to a 6 V battery in series with a resistance of 5$\Omega$. Determine the emf of the primary cell which gives a balance point at 40 cm.
CBSE DELHI - SET 1 2014
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Current flowing in Potentiometer wire,
$\text{I} = \frac{\text{V}}{\text{R} + \text{R}'}$
$ = \frac{6}{10 +5 }\text{A} = 0.4\text{A}$
Potential drop across the potentiometer wire
V = IR
= 0.4 x 10 V = 4.0 V
Potential Gradient k = $\text{V}/ \ell$= 4.0 V/m
$\therefore\text{unknown emf of the cell (E) = K}\ell$
= 4.0 x 0.4 V
= 1.6 V.
$\text{I} = \frac{\text{V}}{\text{R} + \text{R}'}$
$ = \frac{6}{10 +5 }\text{A} = 0.4\text{A}$
Potential drop across the potentiometer wire
V = IR
= 0.4 x 10 V = 4.0 V
Potential Gradient k = $\text{V}/ \ell$= 4.0 V/m
$\therefore\text{unknown emf of the cell (E) = K}\ell$
= 4.0 x 0.4 V
= 1.6 V.
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