# A resistor with a resistance of 2 Ohm was connected to a battery with an EMF of 3V.

**A resistor with a resistance of 2 Ohm was connected to a battery with an EMF of 3V. The voltage drop across the resistor turned out to be 2V. Determine the short-circuit current.**

EMF = 3 V.

R = 2 ohms.

U = 2 V.

Ikor -?

With a short circuit, there is no external resistance of the circuit R = 0, the current reaches the maximum possible value.

Ohm’s law for a closed circuit is: I = EMF / (R + r), where I is the current in the circuit, EMF is the electromotive force of the current source, R is the external resistance of the circuit, r is the internal resistance of the current source.

Let’s write Ohm’s law for a section of the circuit: I = U / R.

EMF / (R + r) = U / R.

(R + r) * U = EMF * R.

R * U + r * U = EMF * R.

r = (EMF * R – R * U) / U = R * (EMF- U) / U.

r = 2 Ohm * (3 V – 2 V) / 2 V = 1 Ohm.

In case of short circuit: Icor = EMF / r.

Icor = 3 V / 1 Ohm = 3 A.

Answer: with a short circuit, the current will be Icor = 3 A.