$ is voltage.

Given that the transformer steps up the voltage from $$22 \mathrm{~V}$$ to $$220 \mathrm{~V}$$ and the load resistance on the secondary side is $$110 \Omega$$, 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 $$I = \frac{V}{R}$$.

The voltage across the secondary coil (and hence across the load resistance) is $$220 \mathrm{~V}$$, and the load resistance is $$110 \Omega$$. Thus, the current in the secondary circuit is:

$$I_{\text{secondary}} = \frac{220 \mathrm{~V}}{110 \Omega} = 2 \mathrm{~A}$$

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,

$$P_{\text{primary}} = P_{\text{secondary}}$$

Applying the formula for power ($$P = IV$$) in both coils, we have:

$$I_{\text{primary}} \times V_{\text{primary}} = I_{\text{secondary}} \times V_{\text{secondary}}$$

We know $$V_{\text{primary}} = 22 \mathrm{~V}$$, $$V_{\text{secondary}} = 220 \mathrm{~V}$$, and we've calculated $$I_{\text{secondary}} = 2 \mathrm{~A}$$. Plugging these values into the equation, we get:

$$I_{\text{primary}} \times 22 = 2 \times 220$$

Solving for $$I_{\text{primary}}$$:

$$I_{\text{primary}} = \frac{2 \times 220}{22} = \frac{440}{22} = 20 \mathrm{~A}$$

Therefore, the current drawn by the primary coil of the transformer is 20 A.

The correct answer is Option A

20 A.

Current Electricity

Continue this paper

Similar Current Electricity PYQs

More practice from this question