When connecting a resistance of 15 ohms to a battery, the ammeter in the circuit shows 1 A. If the same source

When connecting a resistance of 15 ohms to a battery, the ammeter in the circuit shows 1 A. If the same source is closed with a resistance of 8 ohms, the ammeter will show 1.8 A. Find the EMF of the battery and its internal resistance. Draw a diagram.

In this case, the internal resistance of the source and the load resistance will be connected in series, we write Ohm’s law for such a circuit:

I = U / (R + r), where I is the current strength, U is the EMF of the battery, R is the load resistance, r is the internal resistance of the source. Substituting the data for the first and second cases, we get the system:

U / (15 + r) = 1;

U / (8 + r) = 1.8.

U = (15 + r).

(15 + r) / (8 + r) = 1.8;

15 + r = 1.8r + 14.4;

r = 0.6 / 0.8 = 0.75 ohm.

Then:

U = 1 * (15 + 0.75) = 15.75 V.