The turn ratio of a transformers is given as `2:3`. If the current through the primary coil is `3 A`, thus calculate the current through load resistance

Options

A. `1 A`
B. `4.5` A
C. ` 2 A`
D. `1.5` A

Answer

To solve the problem, we will use the relationship between the turns ratio of a transformer and the currents in the primary and secondary coils. The turns ratio is given as \( \frac{N_s}{N_p} = \frac{2}{3} \), where \( N_s \) is the number of turns in the secondary coil and \( N_p \) is the number of turns in the primary coil. The current in the primary coil \( I_p \) is given as \( 3 \, A \).

### Step-by-Step Solution:

1. **Understand the relationship between turns ratio and current:**
The relationship between the turns ratio and the currents in the primary and secondary coils of a transformer is given by:
\[ \frac{N_s}{N_p} = \frac{I_p}{I_s} \]
where \( I_s \) is the current in the secondary coil.

2. **Substitute the known values:**
From the problem, we have:
\[ \frac{N_s}{N_p} = \frac{2}{3} \quad \text{and} \quad I_p = 3 \, A \]
We can substitute these values into the equation:
\[ \frac{2}{3} = \frac{3}{I_s} \]

3. **Cross-multiply to solve for \( I_s \):**
Cross-multiplying gives us:
\[ 2 \cdot I_s = 3 \cdot 3 \]
Simplifying the right side:
\[ 2 \cdot I_s = 9 \]

4. **Isolate \( I_s \):**
To find \( I_s \), divide both sides by 2:
\[ I_s = \frac{9}{2} = 4.5 \, A \]

5. **Conclusion:**
Therefore, the current through the load resistance (which is the same as the current through the secondary coil) is:
\[ I_s = 4.5 \, A \]

### Final Answer:
The current through the load resistance is \( 4.5 \, A \).

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