# If an emf source of 6 V and an internal resistance of 2 Ohm is connected to an external resistance, then a current

**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.

I2 -?

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.