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