Question

A battery of e.m.f. $2.2V$ and internal resistance $1.2\Omega$ is connected to a potentiometer wire of length $10m$ and resistance $9.8\Omega$. The potential drop per cm of length of the wire is

• A. 0.196 mV
• B. 9.16 m V
• C. 19.6 m V
• D. 1.96 m V

Answer 1

Answer: (D)

1.96 m V

$\Rightarrow E=I[1.2+9.8]$

$\frac{2.2}{11}=I$

$\Rightarrow I=0.2A$

P.D across potentiometer $=0.2\times 9.8$

$=1.96V$

$\frac{P.D}{length}=\frac{1.96}{10}=0.196V/m$

$=\frac{0.196}{100}V/cm$

$=1.96mV/cm$