If a source with an EMF of 6 V and an internal resistance of 2 Ohms 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?
I = E / (R + r)
R1 = E / I1 – r = 6/1 -2 = 4 ohms
R2 = 2 * 4 = 8 ohms
I2 = E / (R2 + r) = 6 / (8 + 2) = 0.6 A
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