9 cells each of emf 3V and internal resistance 1 Ω are connected in 3 rows, each having 3 cells in series. The current flowing in the external resistance of 5 Ω connected across this arrangement of the cells is

1.5 A

The correct answer is Option (3) → 1.5 A

Given:

Each cell: emf = 3 V, internal resistance = 1 Ω

Arrangement: 3 rows, each with 3 cells in series → rows connected in parallel.

For one row:

Total emf of series cells → $E_{row} = 3 + 3 + 3 = 9 \text{ V}$

Total internal resistance of row → $r_{row} = 1 + 1 + 1 = 3 \, \Omega$

Now 3 such rows are connected in parallel:

Equivalent emf of parallel rows → $E_{eq} = 9 \, \text{V}$

Equivalent internal resistance → $\frac{1}{r_{eq}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$

∴ $r_{eq} = 1 \, \Omega$

Total circuit resistance:

$R_{total} = R_{ext} + r_{eq} = 5 + 1 = 6 \, \Omega$

Current in external resistance:

$I = \frac{E_{eq}}{R_{total}} = \frac{9}{6} = 1.5 \, \text{A}$

Final Answer:

$I = 1.5 \, \text{A}$