A power transformer (step up) with an `1:8` turns ratio has `60 Hz,120 V` across the primary, the load in the secondary is `10^(4) Omega`.The current in the secondary is

Options

A. 1.2A
B. 0.96A
C. 12mA
D. 96mA

Answer

To find the current in the secondary coil of the transformer, we can follow these steps:

### Step 1: Understand the transformer turns ratio
The transformer has a turns ratio of 1:8. This means that for every 1 turn in the primary coil, there are 8 turns in the secondary coil.

### Step 2: Use the voltage transformation formula
The voltage transformation in a transformer is given by the formula:
\[ \frac{V_2}{V_1} = \frac{N_2}{N_1} \]
where:
- \(V_2\) is the voltage across the secondary,
- \(V_1\) is the voltage across the primary,
- \(N_2\) is the number of turns in the secondary,
- \(N_1\) is the number of turns in the primary.

Given that \(N_2/N_1 = 8/1\) and \(V_1 = 120 \, V\), we can find \(V_2\):
\[ V_2 = V_1 \times \frac{N_2}{N_1} = 120 \, V \times 8 = 960 \, V \]

### Step 3: Calculate the current in the secondary
We know the load resistance in the secondary is \(R = 10^4 \, \Omega\). Using Ohm's law, the current in the secondary can be calculated as:
\[ I_2 = \frac{V_2}{R} \]
Substituting the values we found:
\[ I_2 = \frac{960 \, V}{10^4 \, \Omega} = \frac{960}{10000} = 0.096 \, A \]

### Step 4: Convert to milliampere
To express the current in milliampere (mA), we multiply by 1000:
\[ I_2 = 0.096 \, A \times 1000 = 96 \, mA \]

### Final Answer
The current in the secondary is \(96 \, mA\).

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