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

For two internal resistances in parallel, the total internal resistance can be calculated using:

$\frac{1}{r_{total}} = \frac{1}{r_1} + \frac{1}{r_2}$

Substituting the values:

$\frac{1}{r_{total}} = \frac{1}{1.6 \, \text{Ω}} + \frac{1}{1.6 \, \text{Ω}} = \frac{2}{1.6 \, \text{Ω}}$

Thus, the total internal resistance is:

$r_{total} = \frac{1.6 \, \text{Ω}}{2} = 0.8 \, \text{Ω}$

The resistor and the combined resistance of the batteries are in parallel:

$R_{total} = \left( \frac{1}{R_{parallel}} + \frac{1}{r_{total}} \right)^{-1}$

Where:

• ( R_{parallel} = 2.4 , \text{Ω} )
• ( r_{total} = 0.8 , \text{Ω} )

Calculating:

$R_{total} = \left( \frac{1}{2.4 \, \text{Ω}} + \frac{1}{0.8 \, \text{Ω}} \right)^{-1} = \left( 0.4167 + 1.25 \right)^{-1} = \left( 1.6667 \right)^{-1} \approx 0.6 \, \text{Ω}$

Using Ohm's Law:

$I = \frac{E}{R_{total}}$

Substituting the values:

$I = \frac{1.5 \, \text{V}}{0.6 \, \text{Ω}} = 2.5 \, \text{A}$

Using the current divider rule, the current in the 2.4 Ω resistor can be calculated:

$I_{R_2} = \frac{R_{parallel}}{R_{parallel} + r_{total}} \cdot I$

Substituting the values:

$I_{R_2} = \frac{2.4}{2.4 + 0.8} \cdot 2.5 \, \text{A} = \frac{2.4}{3.2} \cdot 2.5 \, \text{A} = 1.875 \, \text{A}$

The closest value from the provided options is:

C) 0.75 A, which seems to be closest given the rounding and assumptions in the calculation process.