Question

A battery of emf $2 V$ and negligible internal resistance is connected across a uniform wire of length $10 m$ and resistance $30 Ω$ . The appropriate terminals of a cell of emf $1.5 V$ and internal resistance $1 Ω$ is connected to one end of the wire and the other terminal of the cell is connected through a sensitive galvanometer to a slider on the wire. What is the length of the wire that will be required to produce zero deflection of the galvanometer? How will the balancing length change?

(a) When a coil of resistance $5 Ω$ is placed in series with the accumulator.

(b) The cell of $1.5 V$ is shunted with $5 Ω$ resistor?

Accepted Answer

$(a)$ The principle of the potentiometer is base on potential gradient it means the potential difference across the wire having length, $l$ is,

$V = k l$ ,

here $k$ potential gradient.

Here, $V = 2 V$ and length, $l = 10 m$

$\Rightarrow k$ $= A B = \frac{2}{10} = 0.2 V m^{- 1}$ .
Now,

$V_{AC} = 1.5 V$

$∵ V = k l$

$\Rightarrow (0.2) (A C) = 1.5 V$
$∴ A C = 7.5 m$

Now voltage across the wire $A B$ , using the concept of comparing emf of two secondary cells,
$\frac{V_{AB}}{2} = (\frac{R_{AB}}{R_{AB} + 5}) \Rightarrow V_{AB} = (\frac{30}{30 + 5}) \times 2 = \frac{12}{7} V$

it means the potential gradient, $k' = \frac{V_{AB}}{A B}$
$∴$ Potential gradient process, $k' = \frac{12}{10} V m^{- 1}$
Now, $V_{AC} = 1.5 V$
$∴ (\frac{12}{10}) (A C_{1}) = 1.5$
$∴ A C_{1} = 8.75 m$

$(b) V_{{AC}_{2}} = 1.5 V$
$\Rightarrow k (A C_{2}) = (\frac{5}{5 + 1}) (1.5) \Rightarrow 0.2 \times (A C_{2}) = (\frac{5}{5 + 1}) (1.5)$
$\Rightarrow A C_{2} = 6.25 m$

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