Step 1: Understanding the Concept:
To find the potential difference between two arbitrary points in a circuit branch, we can apply Kirchhoff's Voltage Law (KVL) by traversing the path from the starting point to the end point.
We sum up all the potential drops and gains along the path. Step 2: Key Formula or Approach:
When traversing the circuit in the direction of the current \(I\):
1. The potential change across a resistor \(R\) is \(-IR\) (a drop).
2. The potential change across a battery from the positive terminal to the negative terminal is \(-V\) (a drop).
The governing equation from point A to point B is:
\[ V_A + \sum (\text{Potential Changes}) = V_B \] Step 3: Detailed Explanation:
Let's analyze the given circuit branch traversing from point A to point B.
The current \(I\) is given as \(2\text{ A}\), flowing from A towards B.
We start at point A with an initial potential \(V_A\).
Moving across the first resistor (\(R_1 = 2\ \Omega\)) in the direction of the current results in a potential drop:
\[ \Delta V_{R1} = - I \times R_1 = - (2\text{ A}) \times (2\ \Omega) = -4\text{ V} \]
Next, we cross the battery. We are moving from the longer parallel line (+) to the shorter thicker line (-).
This indicates a potential drop equal to the electromotive force of the battery:
\[ \Delta V_{\text{battery}} = -3\text{ V} \]
Finally, moving across the second resistor (\(R_2 = 1\ \Omega\)) in the direction of the current results in another potential drop:
\[ \Delta V_{R2} = - I \times R_2 = - (2\text{ A}) \times (1\ \Omega) = -2\text{ V} \]
Equating this to the potential at point B (\(V_B\)), we write the full equation:
\[ V_A - 4\text{ V} - 3\text{ V} - 2\text{ V} = V_B \]
\[ V_A - 9\text{ V} = V_B \]
Rearranging the terms to find the required difference \((V_A - V_B)\):
\[ V_A - V_B = 9\text{ V} \] Step 4: Final Answer:
The potential difference is \(9\text{ V}\).
