Question

In a potentiometer wire experiment, the emf of a battery in the primary circuit is $20$ Volt and its internal resistance is $5\Omega$. There is a resistance box (in series with the battery and the potentiometer wire) whose resistance can be varied from $120\Omega$ to $170\Omega$. Resistance of the potentiometer wire is $75\Omega$. The following potential difference can not be measured using this potentiometer

• A. $5\text{V}$
• B. $6\text{V}$
• C. $7\text{V}$
• D. $8\text{V}$

Answer 1

The correct option is D $8\text{V}$


Potential difference across wire is
$V_w=i\times 75...\left(1\right)$
current flow in circuit
$i=\frac{E}{R_{net}}$
$=\frac{20}{5+75+\left(120\text{to}170\right)}$
for maximum current
$i_{max}=\frac{20}{5+75+120}=\frac{1}{10}\text{Amp}.$
$\therefore$ from equation (2)
$V_w=\frac{1}{10}\times 75=7.5\text{Volt}$
Maximum potential that can be measured using secondary circuit across wire is $7.5\text{Volt}$
So, we can not measure $8\text{V}$ through this potentiometer.