Question

The given potentiometer has its wire of resistance $10\Omega$. When the sliding contact is in the middle of the potentiometer wire, the potential drop across $2\Omega$ resistor is :

• A. $10V$
• B. $5V$
• C. $\frac{40}{9}V$
• D. $\frac{40}{11}V$

Answer 1

Answer: (C)

$\frac{20-V_0}{5}+\frac{0-V_0}{5}+\frac{20-V_0}{2}=0$
$4+10=\frac{2V_0}{5}+\frac{V_0}{2}$
$14=\frac{4V_0+5V_0}{10}$
$V_0=\frac{140}{9}$ Volt
Potential difference across $2\Omega$ resistor is $20-V_0$
That is $\left(20-\frac{140}{9}\right)$ Volt
Hence answer is $\left(\frac{40}{9}\right)$ Volt