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 :
Correct Option: , 3
Solution:
$\frac{20-\mathrm{V}_{0}}{5}+\frac{0-\mathrm{V}_{0}}{5}+\frac{20-\mathrm{V}_{0}}{2}=0$
$4+10=\frac{2 \mathrm{~V}_{0}}{5}+\frac{\mathrm{V}_{0}}{2}$
$14=\frac{4 \mathrm{~V}_{0}+5 \mathrm{~V}_{0}}{10}$
$\mathrm{V}_{0}=\frac{140}{9}$ Volt
Potential difference across $2 \Omega$ resistor is $20-\mathrm{V}_{0}$
That is $\left(20-\frac{140}{9}\right)$ Volt
Hence answer is $\left(\frac{40}{9}\right)$ Volt
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