Eight cells each having emf 2 V and the internal resistance of 0.02 Ω are connected in series with an external resistance of 9 Ω. The current flowing through the external resistance is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.75 A
Given:
Number of cells, $n = 8$
EMF of each cell, $\mathcal{E} = 2 \ \text{V}$
Internal resistance of each cell, $r = 0.02 \ \Omega$
External resistance, $R = 9 \ \Omega$
Total EMF: $\mathcal{E}_\text{total} = n \mathcal{E} = 8 \cdot 2 = 16 \ \text{V}$
Total internal resistance: $r_\text{total} = n r = 8 \cdot 0.02 = 0.16 \ \Omega$
Current: $I = \frac{\mathcal{E}_\text{total}}{R + r_\text{total}} = \frac{16}{9 + 0.16} = \frac{16}{9.16} \approx 1.746 \ \text{A}$
Current through external resistance: $I \approx 1.75 \ \text{A}$