Question

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

• A. 500$\sqrt{2}$
• B. 10$\sqrt{2}$
• C. 50$\sqrt{2}$
• D. 20$\sqrt{2}$

Answer 1

Answer: (A)

500$\sqrt{2}$

Answer 2

To find the peak value of voltage in the secondary winding, we first determine the RMS voltage in the secondary winding using the transformer's turn ratio and then convert it to the peak value.

Given: Turn ratio ($\frac{N_s}{N_p}$) = 5 RMS value of potential difference on primary ($V_{p_{rms}}$) = 100 V Efficiency of the transformer = 100% (This implies an ideal transformer)

For an ideal transformer, the ratio of secondary voltage to primary voltage is equal to the turn ratio: $\frac{V_{s_{rms}}}{V_{p_{rms}}} = \frac{N_s}{N_p}$

Substitute the given values: $\frac{V_{s_{rms}}}{100 \, \text{V}} = 5$

Solving for $V_{s_{rms}}$: $V_{s_{rms}} = 5 \times 100 \, \text{V}$ $V_{s_{rms}} = 500 \, \text{V}$

Now, to find the peak value of voltage in the secondary winding ($V_{s_{peak}}$), we use the relationship between RMS value and peak value for a sinusoidal AC voltage: $V_{peak} = V_{rms} \times \sqrt{2}$

Therefore, for the secondary winding: $V_{s_{peak}} = V_{s_{rms}} \times \sqrt{2}$ $V_{s_{peak}} = 500 \, \text{V} \times \sqrt{2}$ $V_{s_{peak}} = 500\sqrt{2} \, \text{V}$

The frequency (50 Hz) and the resistance of the secondary winding (500 Ω) are not required for this specific calculation, as we are looking for the peak voltage and the transformer is ideal.