Question

A 12V, 60W lamp is connected to the secondary of a step down transformer, whose primary is connected to ac mains of 220V. Assuming the transformer to be ideal, what is the current in the primary winding?

• A. 0.37 A
• B. 0.27 A
• C. 2.7 A
• D. 3.7 A

Answer 1

The Correct Option is B

An ideal transformer conserves power, so the power on the primary side is equal to the power on the secondary side. The power on the secondary side of the transformer can be calculated using the formula for electrical power:

P = VI 

where P is power, V is voltage, and I is current. Given values for the lamp are:

• Voltage (V) = 12V
• Power (P) = 60W

From the formula, we can find the current on the secondary side (I_s):

I_s = P/V

Substituting the given values:

I_s = 60W / 12V = 5A

Using the transformer equation and the fact that power is conserved:

• Power on primary side P_p = Power on secondary side P_s
• P_p = V_p × I_p
• P_s = V_s × I_s

Since P_p = P_s, we have:

V_p × I_p = V_s × I_s

Given, primary voltage V_p = 220V. Substitute for known values:

220V × I_p = 12V × 5A

I_p = (12V × 5A) / 220V

I_p = 60 / 220

I_p ≈ 0.27A

Therefore, the current in the primary winding is approximately 0.27 A.

Answer 2

\(V_sI_s=V_pI_p\) (ideal transformer)

\(\Rightarrow P_{out}=P_{in}\)

\(60=220\times I_p\)

\(I_p=\frac{60}{220}=0.27A\)

So, the correct answer is option (B): 0.27A