(13 Q:) . Consider a circuit with two batteries and two resistors, all in series

Question

(13 Q:) . Consider a circuit with two batteries and two resistors, all in series. The two batteries have EMFs of 12.0 V and 3.0 V, respectively. They are placed in the circuit in such a way that their EMFs oppose each other. Both batteries have an internal resistance of 0.6 ohms. The two resistors have resistances of 1.6 and 7.7 ohms, respectively. The connecting wire in the circuit has negligible resistance. The layout of the components, going around the loop from a point P0 is as follows: Point P0 is connected to the negative terminal of the 12-V battery; the positive terminal of the 12-V battery is connected to one side of the 1.6-ohm resistor; the other side of the 1.6-ohm resistor is connected to the positive terminal of the 3-V battery; the negative terminal of the 3-V battery is connected to one side of the 7.7-ohm resistor; and the other side of the 7.7-ohm resistor is connected back to P0. How many amperes of current are flowing in the circuit?

Explanation / Answer

i = (12 - 3)/(0.6 + 0.6 + 1.6 + 7.7) = 0.857 A

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