Step 1: Understanding the Concept:
The circuit has three branches between B and D: the branch through A, the branch through C, and the 2 \(\Omega\) resistor.
We use Kirchhoff's two rules. The junction rule says the total current into a node is zero. The loop rule says the net potential change around a closed loop is zero.
The potential difference \(V_B - V_D\) is the potential drop across the 2 \(\Omega\) resistor, or the same value along any other path.
Step 2: Assign the currents:
Let \(I_1\) flow from D through A to B, through the 2 V and 1 V cells.
Let \(I_3\) flow from B to D through the 2 \(\Omega\) resistor.
Let \(I_2\) flow from B through C to D, through the 4 V and 3 V cells.
Junction rule at B: \(I_1 = I_2 + I_3\).
Step 3: Loop rule for loop D-A-B-D:
Go from D to A. The 2 V cell is crossed from its minus to its plus side, so we gain 2 V. Its 1 \(\Omega\) gives a drop \(I_1\).
Go from A to B. The 1 V cell is crossed from plus to minus, so we lose 1 V. Its 2 \(\Omega\) gives a drop \(2I_1\).
Go from B to D along \(I_3\). The 2 \(\Omega\) gives a drop \(2I_3\).
\[ 2 - I_1 - 1 - 2I_1 - 2I_3 = 0 \quad \Rightarrow \quad 3I_1 + 2I_3 = 1 \]
Step 4: Loop rule for loop B-C-D-B:
Go from B to C. The 4 V cell is crossed from minus to plus, so we gain 4 V. Its 3 \(\Omega\) gives a drop \(3I_2\).
Go from C to D. The 3 V cell is crossed from plus to minus, so we lose 3 V. Its 2 \(\Omega\) gives a drop \(2I_2\).
Go from D to B against \(I_3\). The 2 \(\Omega\) now gives a gain \(2I_3\).
\[ 4 - 3I_2 - 3 - 2I_2 + 2I_3 = 0 \quad \Rightarrow \quad 5I_2 - 2I_3 = 1 \]
Step 5: Solve the three equations:
From the two loop equations: \(I_1 = \dfrac{1 - 2I_3}{3}\) and \(I_2 = \dfrac{1 + 2I_3}{5}\).
Put these into \(I_1 = I_2 + I_3\) and multiply by 15:
\[ 5(1 - 2I_3) = 3(1 + 2I_3) + 15I_3 \quad \Rightarrow \quad 2 = 31\,I_3 \]
\[ I_3 = \frac{2}{31}\ \text{A}, \quad I_1 = \frac{9}{31}\ \text{A}, \quad I_2 = \frac{7}{31}\ \text{A} \]
All three currents are positive, so the assumed directions are right.
Step 6: Find the potential difference:
The current \(I_3\) flows from B to D through the 2 \(\Omega\) resistor, so B is at the higher potential.
\[ V_B - V_D = I_3 \times 2 = \frac{2}{31} \times 2 = \frac{4}{31}\ \text{V} \approx 0.13\ \text{V} \]
Final Answer:
The potential difference between B and D is \(\dfrac{4}{31}\) V, which is about 0.13 V. B is at the higher potential.
\[ \boxed{V_B - V_D = \frac{4}{31}\ \text{V} \approx 0.13\ \text{V}} \]
