Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell.

Solution

Potentiometer can be used to measure the internal resistance of the cell.

A cell of emf E is connected across the resistance box through key K₁.

When key K₁ is opened galvanometer shows deflection at the balancing length l₁.

So, E = k $Error converting from MathML to accessible text.$
If both keys are closed, then balancing point is obtained at length l₂ (l₂ < l₁).
So, V = k $Error converting from MathML to accessible text.$
Now, using the relation,
$straight r space space equals straight R space open parentheses straight E over straight V space minus space 1 close parentheses$
Therefore, we have
$Error converting from MathML to accessible text.$

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Current Electricity

Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell.

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