Question

A step-down transformer has a turns ratio of $20 : 1$. If $8\text{ V}$ is applied across a $0.4\ \Omega$ secondary load, then the primary current will be

• A. $2\text{ A}$
• B. $1\text{ A}$
• C. $0.5\text{ A}$
• D. $4\text{ A}$

Hint:

For an ideal transformer, power is perfectly conserved between the primary and secondary stages ($P_{\text{primary}} = P_{\text{secondary}}$). The secondary power is $P_s = \frac{V^2}{R} = \frac{8^2}{0.4} = \frac{64}{0.4} = 160\text{ W}$. Since it is a $20:1$ step-down transformer, the primary voltage must be $20 \times 8 = 160\text{ V}$. Using $P = VI$, the primary current is simply $\frac{160\text{ W}}{160\text{ V}} = 1\text{ A}$!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A step-down transformer has turns ratio 20:1, secondary voltage 8 V across 0.4 Ω load; find primary current I_p.

Step 2: Key Formula or Approach:
I_s = V_s / R_s. For a transformer, I_p / I_s = N_s / N_p.

Step 3: Detailed Explanation:
I_s = 8/0.4 = 20 A. I_p = 20 × (1/20) = 1 A.

Step 4: Final Answer:
Primary current is 1 A, matching option (B).