Question

Two cells \(\varepsilon_1\) and \(\varepsilon_2\) are connected in opposition to each other as shown in the figure. The cell \(\varepsilon_1\) is of emf \(9\ \text{V}\) and internal resistance \(3\ \Omega\). The cell \(\varepsilon_2\) is of emf \(7\ \text{V}\) and internal resistance \(7\ \Omega\). The potential difference between the points \(A\) and \(B\) is:

• A. \(8.4\ \text{V}\)
• B. \(5.6\ \text{V}\)
• C. \(7.8\ \text{V}\)
• D. \(6.6\ \text{V}\)

Hint:

Remember:

• Opposing cells: \[ E_{\text{net}}=E_1-E_2 \]
• Circuit current: \[ I=\frac{E_{\text{net}}}{r_1+r_2} \]
• Terminal voltage: \[ V=E-Ir \]

Answer 1

The Correct Option is B

Concept: When two cells are connected in opposition: \[ E_{\text{net}} = E_1-E_2 \] Current in the circuit is: \[ I=\frac{E_1-E_2}{r_1+r_2} \]

Step 1: Calculate current in the circuit..; Given: \[ E_1=9\ \text{V}, \qquad r_1=3\ \Omega \] \[ E_2=7\ \text{V}, \qquad r_2=7\ \Omega \] Net emf: \[ E_{\text{net}}=9-7=2\ \text{V} \] Total resistance: \[ R=3+7=10\ \Omega \] Therefore: \[ I=\frac{2}{10} \] \[ I=0.2\ \text{A} \]

Step 2: Find potential difference between \(A\) and \(B\). Potential difference across the second cell: \[ V_{AB}=E_2-Ir_2 \] \[ V_{AB}=7-(0.2)(7) \] \[ V_{AB}=7-1.4 \] \[ V_{AB}=5.6\ \text{V} \] Therefore, the correct answer is: \[ \boxed{5.6\ \text{V}} \]