Question

The resistance of a $10m$ long potentiometer wire is $50\Omega$. It is connected in series with a $3V$ battery and $10\Omega$ resistor. The potential difference between two points separated by distance $40cm$ is equal to ______

• A. $0.02V$
• B. $0.1V$
• C. $0.06V$
• D. $1.2V$

Answer 1

The correct option is B $0.1V$

$I=\frac{3}{60}=\frac{1}{20}A\therefore V_p=\frac{1}{20}\times 50=2.5V10m⟶50\Omega 1m⟶\frac{50}{10}\Omega 1cm⟶\frac{5}{100}\Omega 40cm⟶\frac{5}{100}\times 40\Omega 40cm⟶2\Omega \therefore V=2\Omega \times \frac{1}{20}=0.1V$