Question

in the circuit diagram given below, AB is a uniform wire of resistance $15\Omega$ and
length $1$ m. It is connected to a cell $E_1$ of emf $2$V and negligible internal
resistance and a resistance R The balance point with another cell $E_2$ of emf $75$ mV is found at $30$ cm from end A. Calculate the value of R.

Answer 1

Current drawn from the cell , $E_1=2V$
$I=\frac{E_1}{15+R}=\frac{2}{15+R}$
Potential drop across the wire AB
$V_{AB}=I\times 15=\frac{2\times 15}{15+R}=\frac{30}{15+R}$
Since wire length is 1 m or 100 cm.
So, potential gradient along the wire,
$K=\frac{V_{AB}}{100cm}=\frac{30}{100(15+R)}$
At the balance point
$E_2=L{l}_2$
$75mB=\frac{30}{100(15+R)}\times 30cm$
$75\times 10-3\times \left(100\right)(15+R)=900$
$15+R=\frac{9000}{75}$
$\therefore R=120-15=105ohm$