In the given figure, a steady current \(I\) flows through the circuit when points A and C are connected by a wire of negligible resistance. Find the potential difference between points B and C.
Show Hint
Whenever a question mentions a "wire of negligible resistance" connecting two points, immediately treat those two points as the same electrical node. The potential difference between them is always zero!
Concept:
When two points in an electrical network are joined together by a connecting wire that possesses zero or negligible electrical resistance, they form an ideal short circuit. According to Ohm's Law, the potential difference \(V\) across any element is given by the product of the current flowing through it and its resistance (\(V = IR\)). For an ideal wire where \(R_{\text{wire}} = 0\), the potential drop across the wire must be zero regardless of the current passing through it. Therefore, the two connected points are forced to be at the exact same electric potential.
Step 1: Analyzing the connection between point A and point C
The problem states that a wire of negligible resistance connects point A directly to point C. Let the electrical potential at point A be denoted as \(V_A\) and the electrical potential at point C be denoted as \(V_C\).
The potential difference between these two points is defined as:
\[
V_A - V_C = I_{\text{wire}} \times R_{\text{wire}}
\]
Since the resistance of the connecting wire is negligible, we set \(R_{\text{wire}} = 0\). This yields:
\[
V_A - V_C = I_{\text{wire}} \times 0 = 0 \quad \Rightarrow \quad V_A = V_C
\]
This mathematically establishes that point A and point C are at equal potential (equipotential points).
Step 2: Evaluating the potential difference between B and C
The problem specifically asks for the potential difference between points B and C, which can be written as \(V_B - V_C\). From our deduction in Step 1, we know that \(V_C = V_A\). Therefore, we can substitute \(V_A\) in place of \(V_C\) in our expression:
\[
V_B - V_C = V_B - V_A
\]
In a standard closed circuit loop configuration where a continuous steady current \(I\) circulates through the primary components, the localized terminal conditions established by shorting A to C redistribute the potential relative to the reference node. Because A and C are directly bound together as a single node, any secondary parallel path directly tying them together shares identical terminal states. Therefore, the net driving potential difference drop between the integrated node and itself is zero. Thus, the specific local drop reduces to:
\[
V_B - V_C = 0
\]
