An old car battery that has an emf of e m f1 = 11.3 V and an internal resistance of 57.0 mΩ is connected to an R = 2.00 Ω resistor. In an attempt to recharge the battery, you connect a second battery that has an emf of e m f2 = 12.6 V and an internal resistance of 10.0 mΩ in parallel with the first battery and the resistor with a pair of jumper cables.

a) The diagram of the circuit is in the attached image to this solution.

b) Current in the branches.

Current in the branch of the 2 Ω resistor = 6.20 A

Current in the 11.3 V, 57.0 mΩ, emf source branch = 0.925 A

Current in the 12.6 V, 10.0 mΩ, emf source branch = 5.275 A

Explanation:

For parallel connection of two branches, the voltage in the two branches should be equal. But, 11.3 V ≠ 12.6 V

Hence, the difference in potential difference, causes the battery with larger emf to begin to charge the one with smaller emf. Current from the positive terminal of the battery with larger emf moves to the positive terminal of the battery with smaller emf. It charges the smaller emf source until the emf across the two batteries are the same.

The way to combine their emfs is to convert both emf sources to current sources, combine the current sources and then convert the combined current source into an emf source.

For the 11.3 V and 57.0 mΩ = 0.057 Ω source,

As a current source, I = (V/R) (Ohm's law)

I = (11.3/0.057) = 198.2 A

Current source of current 198.2 A and internal resistance r = 57.0 mΩ

For the 12.6 V and 10.0 mΩ = 0.01 Ω source,

I = (V/R)

I = (12.6/0.01)

I = 1260 A

Current source of current 1260A and internal resistance, 10.0 mΩ

When the current sources are combined,

Current = 198.2 + 1260 = 1458.2 A

Resistance = (0.057//0.01) = (0.057×0.01)/(0.057+0.01)

Resistance = 0.00851 Ω = 8.51 mΩ

Then as a combined voltage source

Voltage = (1458.2×0.00851) = 12.41 V

So, voltage source of voltage, 12.41 V and internal resistance of 8.51 mΩ

The current produced that passes through the 2 Ω resistor can then be calculated from simple circuit formulas

E = I(R + r)

12.41 = I (2 + 0.00851)

I = (12.41/2.00851)

I = 6.20 A

Therefore, current through the 2 Ω resistor = 6.20 A

Current in the branches of the internal resistances will be calculated using current divider rules.

Sum of the internal resistances = 0.057 + 0.01 = 0.067 Ω

Current through the 0.057 Ω resistor = 6.20 × (0.01/0.067) = 0.925 A

Current through the 0.01 Ω resistor = 6.20 × (0.057/0.067) = 5.275 A

Hope this Helps!!!

none, its a machine