To find the current drawn by the primary coil of the transformer, we can use the concept of power conservation in an ideal transformer. The power input to the primary coil is equal to the power output from the secondary coil. The formula for power is
P=IV, where
I is current and
V is voltage.
Given that the transformer steps up the voltage from
22V to
220V and the load resistance on the secondary side is
110Ω, we can first calculate the current in the secondary circuit using Ohm's law,
V=IR, where
V is voltage,
I is current, and
R is resistance. Rearranging this, we get
I=‌.
The voltage across the secondary coil (and hence across the load resistance) is
220V, and the load resistance is
110Ω. Thus, the current in the secondary circuit is:
I‌secondary ‌=‌=2ANow, according to the principle of an ideal transformer (neglecting any losses), the power in the primary coil is equal to the power in the secondary coil. Thus,
P‌primary ‌=P‌secondary ‌Applying the formula for power
(P=IV) in both coils, we have:
I‌primary ‌×V‌primary ‌=I‌secondary ‌×V‌secondary ‌
We know
V‌primary ‌=22V,V‌secondrry ‌=220V, and we've calculated
I‌secondary ‌=2A. Plugging these values into the equation, we get:
I‌primary ‌×22=2×220Solving for
I‌primary ‌ :
I‌primary ‌=‌=‌=20ATherefore, the current drawn by the primary coil of the transformer is
20A.
The correct answer is Option A
20A.