Question

A $6\,volt$ battery is connected to the terminals of a three meter long wire of uniform thickness and resistance of $100\,ohm$ . The difference of potential between two points on the wire separated by a distance of $50\,cm$ will be:
A. $3\,v$
B. $1\,v$
C. $1.5\,v$
D. $2\,v$

Answer 1

Hint: the potential difference is defined as the difference between the energy of the charges between two points of a wire. Here, the potential difference across the wire of length $3\,m$ is given. Now, to calculate the potential difference across the wire of length $50\,cm$ we will first calculate the current in the circuit.

Formula used:
Here, we will use Ohm’s law to calculate the current and the potential difference of the wire, which is given by
$V = IR$
Here, $V$ is the voltage, $I$ is the current in the circuit, and $R$ is the resistance in the circuit.

Complete step by step answer:
Consider a circuit in which resistance of $100\Omega $ is connected to a $3\,m$ long wire which is of a uniform thickness. In this circuit, a battery of $6\,volt$ is connected to the wire.
Therefore, the current in the circuit is given by
$I = \dfrac{V}{R}$
$ \Rightarrow \,I = \dfrac{6}{{100}}$
$ \Rightarrow \,I = 0.06A$
Here, the length of the wire, across which we will calculate the potential drop, is $50\,cm$ .Therefore,

$\displaylines{
  50\,cm = \dfrac{{50}}{{100}} \cr
   = 0.5m \cr} $
Now, the resistance across the wire of length $3\,m$ is $100\Omega $ .
Therefore, the resistance across the wire of length $0.5\,m$ can be calculated as shown below
$R = \dfrac{{100}}{3} \times 0.5$
$ \Rightarrow \,R = \dfrac{{50}}{3}\Omega $
Now, the potential difference across the wire of length $0.5\,m$ can be calculated as shown below
$V = IR$
$ \Rightarrow \,V = 0.06 \times \dfrac{{50}}{3}$
$ \Rightarrow \,V = \dfrac{3}{3}$
$ \Rightarrow \,V = 1\,v$
Therefore, the potential difference between the two points of a wire separated by a distance of $0.5\,m$ is $1\,v$ .

So, the correct answer is “Option B”.

Note:
An alternate way to calculate the potential difference in the wire of length $0.5\,m$ is given by
$\dfrac{{{E_1}}}{{{L_1}}} = \dfrac{{{E_2}}}{{{L_2}}}$
Here, ${E_1}$ is the potential difference across the wire of length, ${L_1} = 3\,m$ and ${E_2}$ is the potential difference across the wire of length ${L_2} = 0.5\,m$ . Here, the potential difference across the wire of the length $3\,m$ is $6\,volt$ . therefore, putting these values in the above equation, we get
$\dfrac{6}{3} = \dfrac{{{E_2}}}{{0.5}}$
$ \Rightarrow \,{E_2} = 1\,v$
Which is the required value of the potential difference.