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:
We are given a step-down transformer with a turns ratio of $\frac{N_p}{N_s} = \frac{20}{1}$. The secondary coil circuit operates at a voltage of $V_s = 8\text{ V}$ across a resistive load of $R_s = 0.4\ \Omega$. We need to compute the current flowing through the primary coil input stage, $I_p$.

Step 2: Key Formula or Approach:
1. First, use Ohm's Law on the secondary output side to find the secondary current ($I_s$): $$I_s = \frac{V_s}{R_s}$$ 2. Next, use the fundamental transformer current relation, which states that current is inversely proportional to the turns ratio: $$\frac{I_p}{I_s} = \frac{N_s}{N_p} \implies I_p = I_s \times \left( \frac{N_s}{N_p} \right)$$

Step 3: Detailed Explanation:
First, calculate the current flowing through the secondary coil stage using Ohm's Law: $$I_s = \frac{8\text{ V}}{0.4\ \Omega} = \frac{80}{4} = 20\text{ A}$$ Now, substitute the secondary current $I_s = 20\text{ A}$ and the inverse turns ratio $\frac{N_s}{N_p} = \frac{1}{20}$ into the current transformation equation: $$I_p = 20 \times \left( \frac{1}{20} \right)$$ The factor of 20 cancels out perfectly: $$I_p = 1\text{ A}$$ This matches option (B).

Step 4: Final Answer:
The primary current in the transformer is $1\text{ A}$, which corresponds to option (B).