Two identical batteries each of emf 1.5 V and internal resistance 1.6 Ω are connected in parallel - AQA - A-Level Physics - Question 28 - 2021 - Paper 1

The total internal resistance ( R_{internal} ) of two identical batteries in parallel is given by:

$R_{internal} = \frac{r}{n} = \frac{1.6 ext{ Ω}}{2} = 0.8 ext{ Ω}$

where r is the internal resistance of one battery and n is the number of batteries.

The total resistance ( R_{total} ) in the circuit when the 2.4 Ω resistor is connected in parallel with the internal resistance is given by:

$\frac{1}{R_{total}} = \frac{1}{R_{internal}} + \frac{1}{R_{load}}$

Substituting the values, we have:

$\frac{1}{R_{total}} = \frac{1}{0.8} + \frac{1}{2.4}$

Calculating the right-hand side gives:

$\frac{1}{R_{total}} = 1.25 + 0.4167 = 1.6667$

Thus,

$R_{total} = \frac{1}{1.6667} \approx 0.6 ext{ Ω}$

Using Ohm's law, the total current ( I_{total} ) provided by the batteries can be calculated as:

$I_{total} = \frac{E_{total}}{R_{total}} = \frac{1.5 ext{ V}}{0.6 ext{ Ω}} = 2.5 ext{ A}$

The current through the 2.4 Ω resistor can be determined using the current division rule:

$I_{R} = \frac{R_{internal}}{R_{internal} + R_{load}} \times I_{total}$

Substituting the known values:

$I_{R} = \frac{0.8}{0.8 + 2.4} \times 2.5 = \frac{0.8}{3.2} \times 2.5 = 0.625 ext{ A}$

This suggests that the correct option is not explicitly given, but rounding down to the nearest available option would yield:

Answer: Option B (0.47 A) is the closest to our calculations, but the computed value approaches 0.625 A.