Two current sources with EMF 40 V and 20 V are connected in series. Determine the internal resistance of the second current source, if the internal resistance of the first current source is 2 Ohm, the resistance of the external circuit is 7 Ohm, the current in the circuit is 2 A.
EMF1 = 40 V.
EMF2 = 20 V.
r1 = 2 ohms.
R = 7 ohms.
I = 2 A.
For a closed circuit, Ohm’s law is valid: I = EMF / (r + R), Where EMF is the electromotive force of the current source, r is the internal resistance of the current source, R is the resistance of the external circuit.
With a series connection of current sources, EMF = EMF1 + EMF2, r = r1 + r2.
I = (EMF1 + EMF2) / (r1 + r2 + R).
r1 + r2 + R = (EMF1 + EMF2) / I.
r2 = (EMF1 + EMF2) / I – r1 – R.
r2 = (40 V + 20 V) / 2 A – 2 ohms – 7 ohms = 21 ohms.
Answer: the internal resistance of the second current source r2 = 21 Ohm.
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