Step 1: Understanding the Concept:
When two voltage cells are connected in an opposition layout (with their identical terminals facing each other, such as positive-to-positive or negative-to-negative), they push electrical current in opposite directions around the circuit loop. The cell with the larger electromotive force ($\text{emf}$) determines the overall direction of current flow. This dominant cell undergoes discharging, while the weaker cell is forced into a charging state.
Step 2: Key Formula or Approach:
1. Find the net effective electromotive force ($E_{\text{net}}$) and total internal resistance ($r_{\text{total}}$) of the closed series loop:
\[ E_{\text{net}} = \varepsilon_1 - \varepsilon_2 \quad (\text{since } \varepsilon_1>\varepsilon_2) \]
\[ r_{\text{total}} = r_1 + r_2 \]
2. Calculate the loop circuit current ($I$) via Ohm's Law:
\[ I = \frac{E_{\text{net}}}{r_{\text{total}}} \]
3. Find the potential difference across terminal points $A$ and $B$ ($V_{AB}$) using either the discharging cell equation or charging cell equation.
Step 3: Detailed Explanation:
Given parameter values:
- $\varepsilon_1 = 9 \text{ V}$, $r_1 = 3 \ \Omega$
- $\varepsilon_2 = 7 \text{ V}$, $r_2 = 7 \ \Omega$
First, determine the net loop voltage and resistance properties:
Because the cells oppose each other and $\varepsilon_1>\varepsilon_2$, the net driving voltage is:
\[ E_{\text{net}} = 9 \text{ V} - 7 \text{ V} = 2 \text{ V} \]
The internal resistances are connected in series along the loop path, so they add up normally:
\[ r_{\text{total}} = r_1 + r_2 = 3 \ \Omega + 7 \ \Omega = 10 \ \Omega \]
Now compute the circulating electrical current ($I$) passing through the circuit:
\[ I = \frac{E_{\text{net}}}{r_{\text{total}}} = \frac{2 \text{ V}}{10 \ \Omega} = 0.2 \text{ A} \]
The current flows out of the positive terminal of $\varepsilon_1$ and enters the positive terminal of $\varepsilon_2$.
Next, calculate the potential difference between points $A$ and $B$ ($V_{AB}$). Points $A$ and $B$ represent the external terminals spanning both cell groups. We can evaluate this terminal value from either side:
- Method A: Analyzing the discharging cell ($\varepsilon_1$):
\[ V_{AB} = \varepsilon_1 - I r_1 \]
\[ V_{AB} = 9 \text{ V} - (0.2 \text{ A} \times 3 \ \Omega) \]
\[ V_{AB} = 9 - 0.6 = 8.4 \text{ V} \]
- Method B: Analyzing the charging cell ($\varepsilon_2$):
Because current is forced backward into its positive terminal, it gains potential across its internal resistance:
\[ V_{AB} = \varepsilon_2 + I r_2 \]
\[ V_{AB} = 7 \text{ V} + (0.2 \text{ A} \times 7 \ \Omega) \]
\[ V_{AB} = 7 + 1.4 = 8.4 \text{ V} \]
Both calculation paths yield an identical result of $8.4\text{ V}$. This matches option (A).
Step 4: Final Answer:
The potential difference between points A and B is 8.4 V.