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

Concept:
• When points A and C are connected by a wire of zero resistance, the potential at A becomes equal to the potential at C ($V_A = V_C$).
• The circuit forms a single closed loop containing two batteries and two resistors.
• We can apply Kirchhoff's Voltage Law (KVL) around this loop to find the steady current, and then use Ohm's law to find the potential difference across specific points.

Step 1: Establish the loop equation
The circuit branch contains two cells. Looking at the standard battery symbols (long line is positive, short thick line is negative): Between A and B: $E_1 = 6\text{V}$, $R_1 = 1\ \Omega$. The positive terminal faces A. Between B and C: $E_2 = 4\text{V}$, $R_2 = 3\ \Omega$. The positive terminal faces B. Let us assume a steady current $I$ flows from left to right (from A to C inside the branch). Applying KVL from A to C: \[ V_A - E_1 - I \cdot R_1 - E_2 - I \cdot R_2 = V_C \]
\[ V_A - 6 - I(1) - 4 - I(3) = V_C \]
\[ V_A - V_C = 10 + 4I \]

Step 2: Calculate the current in the circuit
Since A and C are connected by a wire of negligible resistance, they are at the same potential. \[ V_A = V_C \implies V_A - V_C = 0 \]
Substitute this into the equation: \[ 0 = 10 + 4I \]
\[ 4I = -10 \]
\[ I = -2.5 \text{ A} \]
The negative sign indicates that the actual current flows in the opposite direction, i.e., from C to A through the branch components. So, actual current $I_{actual} = 2.5 \text{ A}$ flowing from right to left (C $\rightarrow$ B $\rightarrow$ A).

Step 3: Calculate the potential difference between B and C
We need to find $V_B - V_C$. We will trace the path from C to B. When moving from C to B, we are moving with the direction of the actual current ($2.5 \text{ A}$). As we cross the $3\ \Omega$ resistor in the direction of current, potential drops. As we cross the $4\text{V}$ battery from the negative terminal to the positive terminal, potential rises. \[ V_C - I_{actual} \cdot R_2 + E_2 = V_B \]
\[ V_C - (2.5)(3) + 4 = V_B \]
\[ V_C - 7.5 + 4 = V_B \]
\[ V_C - 3.5 = V_B \]
Rearranging for $V_B - V_C$: \[ V_B - V_C = -3.5 \text{ V} \]
The magnitude of the potential difference between points B and C is $3.5 \text{ V}$.

Step 4: Conclusion
The magnitude of the potential difference between points B and C is $3.5 \text{ V}$.