Question

The potential difference between nodes \(A\) and \(B\) in a circuit is \(v_A - v_B = 5\) V. The work done in moving a charge of \(1\) coulomb from point \(B\) to \(A\) is ____ J. (answer in integer)

Hint:

Work done in moving a charge q from B to A equals q times the potential difference (v_A - v_B).

Answer 1

Correct Answer: 5

Step 1: Recall the definition of potential difference and work.
The potential difference between two points tells us how much work is needed to move a unit positive charge from one point to the other. If the potential difference between points \(A\) and \(B\) is \(v_A - v_B\), then the work done in moving a charge \(q\) from \(B\) to \(A\) is given by
\[ W = q\,(v_A - v_B) \]
This follows directly from the definition of voltage as work done per unit charge: moving a charge from the lower potential point to the higher potential point requires positive work equal to charge times the potential rise.

Step 2: Identify the known values.
We are given
\[ v_A - v_B = 5 \text{ V} \]
and the charge being moved is
\[ q = 1 \text{ C} \]
The charge moves from point \(B\) to point \(A\), which matches the direction used in the formula above, so we can substitute directly without changing any sign.

Step 3: Substitute the values into the formula.
\[ W = q\,(v_A - v_B) = (1)(5) \]

Step 4: Compute the result.
\[ W = 5 \text{ J} \]

Final Answer:
The work done in moving the charge of \(1\) coulomb from \(B\) to \(A\) is
\[ \boxed{5 \text{ J}} \]