Question

A potentiometer wire of length 1 m has a resistnce 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.

Answer 1

Since,

1m potentiometer has $10\Omega$ of resistance.

$\Rightarrow 40cm$ length would have = $\frac{10}{1}\times 0.4$ $[R=P\frac{L}{A}\Rightarrow R\alpha L]$

$R_1=4\Omega$ resistance

At balance point, current through both driver circuit and connected cell would be same. (I = constant)

Now from onm's Law : $V=1R$ we get,

$\Rightarrow \frac{E}{R_1}=\frac{V}{R_2}$ [E : Emf of primary cell]

$\Rightarrow \frac{E}{4}=\frac{6}{5}$

Gives,

$\Rightarrow E=4.8V$