Seven lead accumulators, each of emf 2 V, and internal resistance 0.07 Ω, are connected in series with an external resistance 9.51 Ω. The current in the circuit is

1.4 A

The correct answer is Option (1) → 1.4 A

Total EMF: $E_\text{total} = 7 \times 2 = 14 \, V$

Total internal resistance: $r_\text{total} = 7 \times 0.07 = 0.49 \, \Omega$

Total resistance in circuit: $R_\text{total} = R_\text{ext} + r_\text{total} = 9.51 + 0.49 = 10 \, \Omega$

Circuit current: $I = \frac{E_\text{total}}{R_\text{total}} = \frac{14}{10} = 1.4 \, A$

Current $I = 1.4$ A