COMEDK 2023 Shift 1 Paper

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=‌

V

R

.
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 ‌=‌

220V

110Ω

=2A
Now, 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:
We know V‌primary ‌=22V,V‌secondrry ‌=220V, and we've calculated I‌secondary ‌=2A. Plugging these values into the equation, we get:
I‌primary ‌×22=2×220
Solving for I‌primary ‌ :
I‌primary ‌=‌

2×220

22

=‌

440

22

=20A
Therefore, the current drawn by the primary coil of the transformer is 20A.
The correct answer is Option A
20A.