Question

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.

Hint:

Always clearly define your assumed current direction. A negative result just means the current physically flows opposite to your assumption. Keep the sign consistent when calculating voltage drops.

Answer 1

Step 1: Read the figure:
Cell E1 = 6 V with internal series resistor 1 \(\Omega\) lies between A and B. Cell E2 = 4 V with 3 \(\Omega\) lies between B and C.
In the printed figure the longer plate of E1 is on the A side, so the positive terminal of E1 faces A. The longer plate of E2 is on the C side, so the positive terminal of E2 faces C.
The arrows show the current I flowing from C through B towards A.
A and C are joined by a wire of negligible resistance, so the branch A-B-C becomes a closed loop and \(V_A = V_C\).

Step 2: Find the current in the loop:
Take the loop direction C to B to A, the same as the arrows.
E2 is crossed from its positive plate to its negative plate, so it acts against this direction: \(-4\ \text{V}\).
E1 is crossed from its negative plate to its positive plate, so it acts along this direction: \(+6\ \text{V}\).
Net emf along the loop = \(6 - 4 = 2\ \text{V}\). Total resistance = \(1 + 3 = 4\ \Omega\).
\[ I = \frac{E_1 - E_2}{R_1 + R_2} = \frac{6 - 4}{1 + 3} = 0.5\ \text{A} \] The positive value agrees with the arrows in the figure. Current flows from C to A through the branch.

Step 3: Potential difference between B and C:
Move from C to B along the current.
Crossing E2 from its positive to its negative plate lowers the potential by 4 V.
Crossing the 3 \(\Omega\) resistor along the current lowers it by \(IR_2 = 0.5 \times 3 = 1.5\ \text{V}\).
\[ V_B - V_C = -E_2 - I R_2 = -4 - 1.5 = -5.5\ \text{V} \] So C is at the higher potential.
Check from the other side: from B to A, E1 raises the potential by 6 V and the 1 \(\Omega\) resistor lowers it by 0.5 V, so \(V_A - V_B = +5.5\ \text{V}\). Since \(V_A = V_C\), \(V_C - V_B = 5.5\ \text{V}\). Both routes agree.

Final Answer:
The potential difference between B and C is \(5.5\ \text{V}\) (point C is at the higher potential; \(V_B - V_C = -5.5\ \text{V}\)).