In a network with a voltage of 220 V, two electric lamps with a resistance of 200 Ohm

In a network with a voltage of 220 V, two electric lamps with a resistance of 200 Ohm each are connected in parallel. Determine the strength of the current passing through each lamp.

Since the lamps are connected in parallel, the total resistance of the circuit will be equal to:
1 / (1/200 + 1/200) = 1 / (2/200) = 200/2 = 100 ohms.
Then the total current flowing through the circuit will be equal to:
220: 100 = 2.2 A.
Since the resistance of the lamps is the same, the total current will be divided equally between them:
2.2: 2 = 1.1 A.