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Standard XII

Physics

Potentiometer

In a potentio...

Question

In a potentiometer experiment, the emf of two cells joined in series is -balanced at 6 m length of wire. when the cell of smaller emf is connected by reversing the terminal , the balancing length obtained in 2m. The ratio of the emf of the emf of the cells will be:-

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Solution

$\frac{Δ V}{L} . (6) = E_{1} + E_{2}$ --------------- $(1)$
$\frac{Δ V}{L} (2) = E_{1} - E_{2}$ ---------------- $(2)$
$\Rightarrow 3 = \frac{E_{1} + E_{2}}{E_{1} - E_{2}}$
$\Rightarrow 3 E_{1} - 3 E_{2} = E_{1} + E_{2}$
$\Rightarrow 2 E_{1} = 4 E_{2}$
$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{1}{2}$
$∴$ ratio is $\frac{1}{2} .$

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In a potentiometer experiment, the emf of two cells joined in series is -balanced at 6 m length of wire. when the cell of smaller emf is connected by reversing the terminal , the balancing length obtained in 2m. The ratio of the emf of the emf of the cells will be:-

ΔVL.(6)=E1+E2---------------(1)ΔVL(2)=E1−E2----------------(2)⇒3=E1+E2E1−E2⇒3E1−3E2=E1+E2⇒2E1=4E2⇒E1E2=12∴ ratio is 12.