If a transformer have turn tatio 5, frequency 50 Hz root mean square value of potential difference on primary 100 volts and the resistance of the secondary winding is 500 ` Omega` then the peak value of voltage in secondary winding will be (the efficiency of the transformer is hundred percent)

Options

A. `500 sqrt2`
B. `10 sqrt 2`
C. `50 sqrt2`
D. `20 sqrt2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Transformer Turn Ratio
The turn ratio (n) of a transformer is defined as the ratio of the number of turns in the secondary coil (Ns) to the number of turns in the primary coil (Np). Given that the turn ratio is 5, we can express this as:
\[ \frac{N_s}{N_p} = 5 \]

### Step 2: Use the Transformer Voltage Relationship
The relationship between the primary voltage (V1) and the secondary voltage (V2) in a transformer is given by:
\[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \]
Given that the root mean square (RMS) value of the primary voltage (V1) is 100 volts, we can express this as:
\[ V_p = 100 \text{ volts} \]

### Step 3: Calculate the RMS Value of the Secondary Voltage
Using the turn ratio, we can find the RMS value of the secondary voltage (Vs):
\[ \frac{100}{V_s} = \frac{1}{5} \]
From this, we can solve for Vs:
\[ V_s = 100 \times 5 = 500 \text{ volts} \]

### Step 4: Convert RMS Value to Peak Value
To find the peak value of the secondary voltage (V_peak), we use the relationship between RMS and peak values:
\[ V_{rms} = \frac{V_{peak}}{\sqrt{2}} \]
Thus, we can express the peak value as:
\[ V_{peak} = V_{rms} \times \sqrt{2} \]
Substituting the RMS value of the secondary voltage:
\[ V_{peak} = 500 \times \sqrt{2} \]

### Final Answer
The peak value of the voltage in the secondary winding is:
\[ V_{peak} = 500\sqrt{2} \text{ volts} \]

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