A source of electrical energy, the EMF of which is 4 volts and an internal resistance of 1 ohm, a current of 5 amperes flows.

A source of electrical energy, the EMF of which is 4 volts and an internal resistance of 1 ohm, a current of 5 amperes flows. Determine the readings of the voltmeter connected to the terminals of this source.

Determine the voltage drop across the internal resistance of the source.
According to Ohm’s law
U = I * Rv;
U = 5 * 1 = 5V,
which is not possible, since this value is greater than EMF 4 V.

The total resistance of the circuit is the sum of the internal resistance of the source and the resistance of the load.
R = Rv + Rn.

Determine the maximum possible current
I = E / R = E / (Rv + Rn).

The maximum current will be at short circuit at Rн = 0
I = 4 / (1 + 0) = 4 A.

This means that even with a short circuit, the current cannot be more than 4 A, which does not correspond to the conditions of the problem.

Answer: the conditions of the problem are not correct, there is no solution.