# The transformer increases the voltage from 220V to 3000V. A current of 0.1A flows in the secondary winding.

The transformer increases the voltage from 220V to 3000V. A current of 0.1A flows in the secondary winding. Determine the current in the primary winding, if the efficiency of the transformer is 96%.

U1 = 220 V.
U2 = 3000 V.
I2 = 0.1 A.
Efficiency = 96%.
I1 -?
The efficiency of the transformer is determined by the formula: efficiency = N2 * 100% / N1, where N2 is the power that the transformer gives, N1 is the power that it consumes.
The power of the electric current N is determined by the formula: N = I * U, where I is the current in the winding, U is the voltage in the winding.
N1 = I1 * U1, N2 = I2 * U2.
Efficiency = I2 * U2 * 100% / I1 * U1.
I1 = I2 * U2 * 100% / efficiency * U1.
I1 = 0.1 A * 3000 V * 100% / 96% * 220 V = 1.42 A.
Answer: the current in the primary winding is I1 = 1.42 A.

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