If an emf source of 6 V and an internal resistance of 2 Ohm is connected to an external resistance, then a current of 1 A appears in the circuit. What current will flow in the circuit if the external resistance is doubled?
EMF = 6 V.
r = 2 ohms.
R2 = 2 * R1.
I1 = 1 A.
According to Ohm’s law for a closed loop, the current I in the circuit is directly proportional to the electromotive force of the current source EMF and is inversely proportional to the total resistance of the circuit: I = EMF / (r + R).
I2 = EMF / (r + R2) = EMF / (r + 2 * R1).
Let’s write Ohm’s law for the first case: I1 = EMF / (r + R1).
r + R1 = EMF / I1.
R1 = EMF / I1 – r.
I2 = EMF / (r + 2 * (EMF / I1 – r)).
I2 = 6 V / (2 Ohm + 2 * (6 V / 1 A – 2 Ohm)) = 0.6 A.
Answer: the current strength in the second case is I2 = 0.6 A.
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