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)
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Work done in moving a charge q from B to A equals q times the potential difference (v_A - v_B).
Step 1: Think of potential as stored energy per unit charge.
Every point in a circuit has an electric potential, which is the potential energy a unit positive charge would have if placed there. So a charge $q$ sitting at a point with potential $v$ carries potential energy $qv$.
Step 2: Write the energy at each point for our charge.
For our charge $q=1$ C, the potential energy at point $A$ is $q\,v_A$ and at point $B$ is $q\,v_B$.
Step 3: Use the work-energy link.
The work done by an external agent in moving the charge from $B$ to $A$ equals the gain in potential energy, since the charge moves at constant speed with no change in kinetic energy:
\[ W = q\,v_A - q\,v_B = q\,(v_A - v_B) \]
Step 4: Plug in the numbers.
We are told $v_A - v_B = 5$ V and $q = 1$ C, so
\[ W = (1)(5) = 5 \text{ J} \]
Step 5: Sanity check the sign.
Since $A$ is at the higher potential, moving a positive charge from $B$ up to $A$ takes positive work, which matches our positive result.
\[ \boxed{5 \text{ J}} \]
