Question

In the circuit shown below the cells E₁ and E₂ have emf’s 4 V and 8 V and internal resistance 0.5 ohm and $1ohm$respectively. Then the potential difference across cell E₁ and E₂ will be

• A. 3.75 V, 7.5 V
• B. 4.25 V, 7.5 V
• C. 3.75 V, 3.5 V
• D. 4.25V, 4.25 V

Answer 1

Answer: (B)

4.25 V, 7.5 V

Answer 2

In the given circuit diagram external resistance $R = \frac{3 \times 6}{3 + 6} + 4.5 = 6.5\Omega$. Hence main current through the circuit $i = \frac{E_{2} - E_{1}}{R + r_{eq}} = \frac{8 - 4}{6.5 + 0.5 + 0.5} = \frac{1}{2}amp.$

Cell 1 is charging so from it’s emf equation E₁ = V₁ – ir₁ ⇒ $4 = V_{1} - \frac{1}{2} \times 0.5$ ⇒ V₁ = 4.25 volt

Cell 2 is discharging so from it’s emf equation E₂ = V₂ + ir₂ ⇒ $8 = V_{2} + \frac{1}{2} \times 1$ ⇒ V₂ = 7.5 volt