Given data

• Two identical batteries, each with internal resistance $r = 1\,\Omega$
• External resistor = $R$
• Heat produced per unit time in $R$:
  • In series combination: $J_1$
  • In parallel combination: $J_2$
• Given: $J_1 = 2.25\,J_2$

Power dissipated in the external resistor The rate of heat produced in $R$ is the power in $R$: $P = I^2 R$

Case 1: Batteries in series Let emf of each battery be $E$.

For series combination:

• Equivalent emf: $2E$
• Equivalent internal resistance: $2r = 2\,\Omega$

Current through $R$: $I_1 = \frac{2E}{R+2}$

Hence, $J_1 = I_1^2 R = \left(\frac{2E}{R+2}\right)^2 R$ $J_1 = \frac{4E^2R}{(R+2)^2}$

Case 2: Batteries in parallel For identical batteries in parallel:

• Equivalent emf: $E$
• Equivalent internal resistance: $r_{\text{eq}} = \frac{r}{2} = \frac{1}{2}\,\Omega$

Current through $R$: $I_2 = \frac{E}{R+\frac{1}{2}}$

Hence, $J_2 = I_2^2 R = \left(\frac{E}{R+\frac{1}{2}}\right)^2 R$ $J_2 = \frac{E^2R}{\left(R+\frac{1}{2}\right)^2}$

Use the given ratio $J_1 = 2.25\,J_2 = \frac{9}{4}J_2$

So, $\frac{J_1}{J_2} = \frac{9}{4}$

Substitute expressions: $\frac{\frac{4E^2R}{(R+2)^2}}{\frac{E^2R}{\left(R+\frac{1}{2}\right)^2}} = \frac{9}{4}$

Cancel $E^2R$: $4\cdot \frac{\left(R+\frac{1}{2}\right)^2}{(R+2)^2} = \frac{9}{4}$

Solve the equation $\frac{\left(R+\frac{1}{2}\right)^2}{(R+2)^2} = \frac{9}{16}$

Taking positive square root: $\frac{R+\frac{1}{2}}{R+2} = \frac{3}{4}$

Cross-multiply: $4\left(R+\frac{1}{2}\right) = 3(R+2)$ $4R+2 = 3R+6$ $R = 4\,\Omega$

Final answer $\boxed{4}$

Comparison with stored answer Stored correct answer = $4$

Our derived answer matches the stored answer.