Question

The resistance of a 10m long potentiometer wire is $20\Omega $. It is connected in series with a 3V battery and a $10\Omega $ resistor. What is the potential difference between two points separated by a distance of 30cm?
(A) $0.02V$
(B) $0.06V$
(C) $0.1V$
(D) $1.2V$

Answer 1

Hint
A potentiometer wire has a potential gradient across it. Hence, the potential difference across any two points on the wire is different from the total potential, but the current through the wire remains constant.
$V = IR$; Where, $V$ is the potential difference across two points, $R$ is the resistance in between and $I$ is the current flow

Complete step by step answer
In the given circuit, we have a potentiometer wire joined in series with a resistor across a battery. The specifications include:
Length of the potentiometer wire $L = 10m$
Resistance of the potentiometer wire ${R_w} = 20\Omega $
Resistor in series $10\Omega $
Battery potential $V = 3V$
We know that by Ohm’s Law,
$V = IR$
Rearranging it in terms of $I$, we get:
$I = \dfrac{V}{R}$
This current flows through all of the circuit. Hence, we need to find the total resistance. As the potentiometer wire and the resistor are in series we get the equivalent resistance using simple addition:
${R_{eq}} = {R_w} + R$
Putting the given values gives us:
$
  {R_{eq}} = 20 + 10 \\
   \Rightarrow {R_{eq}} = 30\Omega \\
 $
We put this value in the Ohm’s Law to find the current:
$
  I = \dfrac{3}{{30}} \\
   \Rightarrow I = 0.1A \\
 $
Hence, the total current is $0.1$ Amperes.
We are asked to calculate the potential difference between two points on the wire. We already know the current. To find the resistance of that small section of the wire, we use proportionality as:
10m of wire $ \Rightarrow 20\Omega $ resistance
30cm of wire$ \Rightarrow x\Omega $ resistance
By cross-multiplying, we have:
$x = \dfrac{{20 \times 30cm}}{{10m}}$
Converting cm to m:
$
  x = \dfrac{{20 \times 30}}{{10 \times 100}} \\
   \Rightarrow x = \dfrac{6}{{10}} \\
 $
Hence, resistance on the 30cm portion of the wire is $0.6\Omega $
Again, using Ohm’s Law:
$
  V = 0.1 \times 0.6 \\
   \Rightarrow V = 0.06V \\
 $ [current multiplied by resistance]
$\therefore $ The answer is option B: $0.06V$

Note
Although the same current flows in all of an electric circuit (except for when resistors are connected in parallel), the voltage varies at every point. This is because it depends on both the current and the resistance. An electric circuit seems to exhibit different voltage across resistances at different points.