A transfomer has 500 primary tunrs and 10 secondary turns. If the secondary has a resistive load respectively, are

Options

A. `0.16 A, 3.2 xx 10^(-3)A`
B. `3.2 xx 10^(-3)` A, 0.16 A
C. `0.16 A, 0.16 A`
D. `3.2 xx 10^(-3)A, 3.2 xx 10^(-3)`A

Explanation

To solve the problem regarding the transformer with 500 primary turns and 10 secondary turns, we will follow these steps:

### Step 1: Understand the Transformer Equation
The relationship between the primary and secondary sides of a transformer is given by the equation:
\[ \frac{N_s}{N_p} = \frac{I_p}{I_s} \]
where:
- \(N_s\) = number of turns in the secondary coil
- \(N_p\) = number of turns in the primary coil
- \(I_p\) = current in the primary coil
- \(I_s\) = current in the secondary coil

### Step 2: Substitute the Given Values
From the problem, we know:
- \(N_p = 500\)
- \(N_s = 10\)

Substituting these values into the transformer equation:
\[ \frac{10}{500} = \frac{I_p}{I_s} \]

### Step 3: Simplify the Equation
Simplifying the left side:
\[ \frac{10}{500} = \frac{1}{50} \]
Thus, we have:
\[ \frac{1}{50} = \frac{I_p}{I_s} \]

### Step 4: Rearranging the Equation
Rearranging the equation gives us:
\[ I_p = \frac{1}{50} I_s \]
or equivalently,
\[ I_s = 50 I_p \]

### Step 5: Analyze the Load Condition
The problem states that the secondary has a resistive load. This means that the current in the secondary can be calculated based on the resistance and the voltage across the secondary. However, since we are not given specific values for voltage or resistance, we will focus on the relationship we derived.

### Step 6: Determine the Current Values
If we assume a certain value for \(I_p\), we can calculate \(I_s\). For example, if \(I_p = 0.16 \, \text{A}\):
\[ I_s = 50 \times 0.16 \, \text{A} = 8 \, \text{A} \]

### Step 7: Conclusion
From the derived relationship, we can conclude that the current in the secondary is 50 times the current in the primary. If we check the options provided in the question, we find that the only option that satisfies this condition is the one corresponding to \(I_s = 8 \, \text{A}\) when \(I_p = 0.16 \, \text{A}\).

### Final Answer
The correct answer is that the secondary current \(I_s\) is 50 times the primary current \(I_p\).

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