Consider a circuit with two batteries and two resistors, all in series. The two

Question

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 and 9.3 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-ohm resistor; the other side of the 1-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 9.3-ohm resistor; and the other side of the 9.3-ohm resistor is connected back to P0. At how much higher potential (in volts) is the positive terminal of the 3-V battery than the point P0 (the negative terminal of the 12-V battery)?

Explanation / Answer

Let

e1 = 12 volts

e2 = 3 volts

r1= r2 = 0.5 ohms

R1 = 1.4 ohms

R2 = 5.5 ohms

Let I is the current in the ckt.

Apply, Kirchoff's volatge in the loop,

e1 - I*R1 - e2 - I*r1 - I*R2 - I*r2 = 0

e1 - e2 = I*(R1 + R2 + r1 + R2)

I = (e1 - e2)/(R1 + R2 + r1 + r2)

= (12 - 3)/(1.4 + 5.5 + 0.5 + 0.5)

= 1.139 A <<<<<<<-------------Answer

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