Thirty-six cells of 1.5 V, each with an internal resistance of 0.50 Ω, drive a heater of 2.0 Ω…

Step 1. The rows. $m = \sqrt{\dfrac{Nr}{R}} = \sqrt{\dfrac{36 \times 0.50}{2.0}} = \sqrt{9} = 3$.

Step 2. The row. $k = \dfrac{36}{3} = 12$ cells in each row.

Step 3. The current. The group has $E = 12 \times 1.5\ \text{V} = 18\ \text{V}$ and $r = \dfrac{12 \times 0.50\ \Omega}{3} = 2.0\ \Omega$. So $I = \dfrac{18\ \text{V}}{2.0\ \Omega + 2.0\ \Omega} = 4.5\ \text{A}$.

Step 4. All in series. $E = 54\ \text{V}$ and $r = 18\ \Omega$, so $I = \dfrac{54\ \text{V}}{20\ \Omega} = 2.7\ \text{A}$.