Resistor Value for Equal Power Dissipation in Series and Parallel Battery Combinations

To find the value of the resistor $R$, we analyze the rate of heat produced (power dissipated) in $R$ for both series and parallel combinations of the two identical batteries.

Let each battery have an electromotive force (EMF) of $E$ and an internal resistance of $r = 1\,\Omega$.

Case 1: Batteries in Series

When the two identical batteries are connected in series across the resistor $R$:

• Equivalent EMF, $E_{\text{eq1}} = 2E$
• Equivalent internal resistance, $r_{\text{eq1}} = 2r = 2 \times 1\,\Omega = 2\,\Omega$

The current $I_1$ flowing through the resistor $R$ is given by:
$I_1 = \frac{E_{\text{eq1}}}{R + r_{\text{eq1}}} = \frac{2E}{R + 2}$

The rate of heat produced in $R$ is:
$J_1 = I_1^2 R = \left(\frac{2E}{R + 2}\right)^2 R = \frac{4E^2 R}{(R + 2)^2}$

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Case 2: Batteries in Parallel

When the two identical batteries are connected in parallel across the resistor $R$:

• Equivalent EMF, $E_{\text{eq2}} = E$
• Equivalent internal resistance, $r_{\text{eq2}} = \frac{r}{2} = \frac{1}{2}\,\Omega = 0.5\,\Omega$

The current $I_2$ flowing through the resistor $R$ is given by:
$I_2 = \frac{E_{\text{eq2}}}{R + r_{\text{eq2}}} = \frac{E}{R + 0.5} = \frac{2E}{2R + 1}$

The rate of heat produced in $R$ is:
$J_2 = I_2^2 R = \left(\frac{2E}{2R + 1}\right)^2 R = \frac{4E^2 R}{(2R + 1)^2}$

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Determining the Resistance $R$

We are given that:
$J_1 = 2.25 J_2 = \frac{9}{4} J_2$

Substituting the expressions for $J_1$ and $J_2$:
$\frac{4E^2 R}{(R + 2)^2} = \frac{9}{4} \cdot \frac{4E^2 R}{(2R + 1)^2}$

Canceling $4E^2 R$ from both sides ($E \neq 0, R > 0$):
$\frac{1}{(R + 2)^2} = \frac{9}{4(2R + 1)^2}$

Taking the square root on both sides (since resistance values are positive):
$\frac{1}{R + 2} = \frac{3}{2(2R + 1)}$

Cross-multiplying to solve for $R$:
$2(2R + 1) = 3(R + 2)$
$4R + 2 = 3R + 6$
$4R - 3R = 6 - 2$
$R = 4\,\Omega$