Two batteries are connected in parallel with an external resistance of [tex]12 \, \Omega[/tex]. The first battery has an electromotive force (emf) of [tex]6 \, V[/tex] and an internal resistance of [tex]0.5 \, \Omega[/tex]. The second battery has an emf of [tex]10 \, V[/tex] and an internal resistance of [tex]1 \, \Omega[/tex]. Find the current through the circuit.

To solve the problem of determining the current through each battery when two batteries are connected in parallel, we need to approach it step by step.

First, let's outline the given values:

Battery 1: Voltage ([tex]V_1[/tex]) = 6V, Internal Resistance ([tex]r_1[/tex]) = 0.5Ω

Battery 2: Voltage ([tex]V_2[/tex]) = 10V, Internal Resistance ([tex]r_2[/tex]) = 1Ω

External Resistance ([tex]R[/tex]): 12Ω

When batteries are in parallel, the combined voltage of the batteries is determined using the principle of a parallel circuit - where the voltage across each component in parallel is the same. However, this scenario involves potential difference because of different voltages, hence Kirchhoff's laws are helpful.

Step 1: Calculate the equivalent internal resistance, [tex]R_{eq}[/tex]:

Given that the batteries are in parallel, the equivalent internal resistance [tex]R_{eq}[/tex] is calculated as follows:

[tex]\frac{1}{R_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{0.5} + \frac{1}{1} = 2 + 1 = 3[/tex]

Thus, [tex]R_{eq} = \frac{1}{3} \approx 0.333 \text{Ω}[/tex]

Step 2: Determine the total current ([tex]I_{total}[/tex]) using an average voltage ([tex]V_{avg}[/tex]):

An effective way is to balance the voltages between the two sources using their internal resistances as weights. Therefore:

[tex]V_{avg} = \frac{V_1 \cdot (1/r_1) + V_2 \cdot (1/r_2)}{(1/r_1) + (1/r_2)}[/tex]

[tex]V_{avg} = \frac{6 \cdot 2 + 10 \cdot 1}{3} = \frac{12 + 10}{3} = \frac{22}{3} \approx 7.33 \text{V}[/tex]

Now, applying Ohm's law to find the current through the external resistance:

[tex]I_{total} = \frac{V_{avg}}{R + R_{eq}} = \frac{7.33}{12 + 0.333} \approx \frac{7.33}{12.333} \approx 0.594 \text{A}[/tex]

Step 3: Determining currents [tex]I_1[/tex] and [tex]I_2[/tex] through each battery:

Using current division principle:

[tex]I_1 = I_{total} \cdot \frac{1/ r_2}{(1/r_1) + (1/r_2)} = 0.594 \cdot \frac{1}{3} \approx 0.198 \text{A}[/tex]

[tex]I_2 = I_{total} \cdot \frac{1/ r_1}{(1/r_1) + (1/r_2)} = 0.594 \cdot \frac{2}{3} \approx 0.396 \text{A}[/tex]

Thus the current through the 6V battery, [tex]I_1[/tex], is approximately 0.198 A, and the current through the 10V battery, [tex]I_2[/tex], is approximately 0.396 A.