A battery of emf 4 V and internal resistance 1 Ω is connected across the wire of resistance of 9.0 Ω. The number of electrons passing through the cross-section of the wire in 4 s is

$10^{19}$

The correct answer is Option (1) → $10^{19}$

Given: EMF $E = 4 \, \text{V}$, internal resistance $r = 1 \, \Omega$, wire resistance $R = 9 \, \Omega$, time $t = 4 \, \text{s}$, electron charge $e = 1.6 \times 10^{-19} \, \text{C}$

Total resistance: $R_\text{total} = R + r = 9 + 1 = 10 \, \Omega$

Current in the circuit:

$I = \frac{E}{R_\text{total}} = \frac{4}{10} = 0.4 \, \text{A}$

Total charge passing through wire in 4 s:

$Q = I t = 0.4 \cdot 4 = 1.6 \, \text{C}$

Number of electrons:

$n = \frac{Q}{e} = \frac{1.6}{1.6 \times 10^{-19}} = 10^{19}$

Number of electrons ≈ 10¹⁹