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) Draw a diagram of the circuit.
(b) Find the current in each branch of the circuit.

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

b) Current in the branches.

Current in the 2 Ω resistor branch = 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:

The first basic underlying principle for parallel connection is that voltage in the two branches are equal. But, 11.3 V ≠ 12.6 V

Hence, the difference in potential difference, causes the battery with bigger emf to begin to charge the one with smaller emf. Current from the positive terminal of the battery with larger emf leave to go to the positive terminal of the battery with smaller emf. It charges it until the end 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 end source.

11.3 V and 57.0 mΩ = 0.057 Ω

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

12.6 V and 10.0 mΩ = 0.01 Ω

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!!!

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