A potentiometer wire of length 1.0 m has a resistance of l5 $\Omega$. It is connected to a 5 V battery in series with a resistance of 5 $\Omega$ Determine the emf of the primary cell which gives a balance point at 60 cm.
CBSE DELHI - SET 2 2014
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$ \text{I}=\frac{V}{\text{R+R}^{'}}$
$=\frac{5}{5+15}\text{A}=0.25\ \text{A}$
Potential drop across the potentiometer wire
V = IR
= 0.25 x 15 V= 3.75 volt
Potential Gradient k = V/ $\ell$ = 3.75 V/ 1.0 m = 3.75 V/m
$\therefore$ unknown emf (E) of the cell = $\text{kl}{'}$
= 3.75 x 0.6 V
= 2.25 volt.
$=\frac{5}{5+15}\text{A}=0.25\ \text{A}$
Potential drop across the potentiometer wire
V = IR
= 0.25 x 15 V= 3.75 volt
Potential Gradient k = V/ $\ell$ = 3.75 V/ 1.0 m = 3.75 V/m
$\therefore$ unknown emf (E) of the cell = $\text{kl}{'}$
= 3.75 x 0.6 V
= 2.25 volt.
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