Question

What does an electric circuit mean? Name a device that helps to maintain a potential difference across a conductor in a circuit. When do we say that the potential difference across a conductor is 1 volt? Calculate the amount of work done in shifting a charge of 2 coulombs from a point A and B having potential $10\,{\text{V}}$ and $5\,{\text{V}}$ respectively.

Answer 1

Recall the basics of an electric circuit like its definition and why potential difference remains same across the circuit. Recall the concept of one volt or definition of one volt. Use the relation between the work done, charge and potential difference to calculate the work done asked in the last part.

Complete step by step answer:
An electric circuit is the closed path along which the electrons and hence current from the electron source flows.The device that help to maintain a potential difference across a conductor in a circuit is a battery.When a work of amount one joule ($1\,{\text{J}}$) is done in moving a unit charge of one coulomb ($1\,{\text{C}}$) from one place to the other place in the conductor, we say that the potential difference across the conductor is one volt ($1\,{\text{V}}$).
From this definition of one volt, we can derive the formula
$1\,{\text{V}} = \dfrac{{1\,{\text{J}}}}{{1\,{\text{C}}}}$
Hence, the formula of potential difference becomes
$V = \dfrac{W}{Q}$ …… (1)
Here, $V$ is the potential difference across the conductor, $W$ is the work done and $Q$ is the charge.We have given that a charge of $2\,{\text{C}}$ is shifted from point A with voltage $10\,{\text{V}}$ to point B with voltage $5\,{\text{V}}$.
$Q = 2\,{\text{C}}$
$\Rightarrow {V_A} = 10\,{\text{V}}$
$\Rightarrow {V_B} = 5\,{\text{V}}$
First, we have to determine the potential difference ${V_{AB}}$ between the points A and B of the conductor.
${V_{AB}} = {V_A} - {V_B}$
Substitute $10\,{\text{V}}$ for ${V_A}$ and $5\,{\text{V}}$ for ${V_B}$ in the above equation.
${V_{AB}} = \left( {10\,{\text{V}}} \right) - \left( {5\,{\text{V}}} \right)$
$\Rightarrow {V_{AB}} = 5\,{\text{V}}$
Hence, the potential difference across points A and B in the conductor is $5\,{\text{V}}$.
Rearrange equation (1) for the work done $W$ in moving the charge between points A and B.
$W = {V_{AB}}Q$
Substitute $5\,{\text{V}}$ for ${V_{AB}}$ and $2\,{\text{C}}$ for $Q$ in the above equation.
$W = \left( {5\,{\text{V}}} \right)\left( {2\,{\text{C}}} \right)$
$\therefore W = 10\,{\text{J}}$

Hence, the work done in shifting the charge from A to point B is $10\,{\text{J}}$.

Note: The students might get confused while solving the last part in the question and take the value of potential difference as the potential difference at point A or point B. But the potential difference across a conductor is the difference between the voltages at two points. So, the correct way to determine the potential difference is to take the difference between two voltages.