Question

The resistance of $10$m long potentiometer wire is $20\Omega$. It is connected in series with a $3$V battery and $10\Omega$ resistor. The potential difference between two points separated by distance $30$cm is equal to _________.

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

Answer 1

The correct option is B $0.06$V

Using Ohm's Law $V=IR$

Current in the circuit = $\frac{V}{R_{eq}}$, And $R_{eq}=(20+10)\Omega$

So, $I=\frac{3}{30}=0.1A$

Given that resistance of $10m$ long potential wire is $20\Omega$

$\Rightarrow$ Resistance of $30cm$ long wire = $\frac{20\times 30}{10\times 100}=0.6\Omega$

Using Ohm's Law again, potential difference between two points = $IR=0.1\times 0.6=0.06V$.

Therefore, B is correct option.