An ideal transformer has `100` turns in the primary and `250` turns in the secondary. The peak value of the ac is `28 V`. The r.m.s. secondary voltage is nearest to

Options

A. 50 V
B. 70 V
C. 100 V
D. 40 V

Answer

To solve the problem step by step, we need to follow the principles of an ideal transformer and the relationship between peak voltage and RMS voltage.

### Step 1: Determine the turns ratio of the transformer
The turns ratio (Ns/Np) can be calculated using the number of turns in the primary (Np) and secondary (Ns) coils.

Given:
- Np = 100 turns (primary)
- Ns = 250 turns (secondary)

**Calculation:**
\[ \text{Turns ratio} = \frac{N_s}{N_p} = \frac{250}{100} = 2.5 \]

### Step 2: Relate the voltages using the turns ratio
For an ideal transformer, the ratio of the secondary voltage (Vs) to the primary voltage (Vp) is equal to the turns ratio.

Using the formula:
\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \]

We know the peak voltage (Vp) is given as 28 V.

**Calculation:**
\[ V_s = V_p \times \frac{N_s}{N_p} = 28 \, \text{V} \times 2.5 = 70 \, \text{V} \]

### Step 3: Calculate the RMS voltage from the peak voltage
The RMS voltage (Vrms) can be calculated from the peak voltage (Vs) using the formula:
\[ V_{rms} = \frac{V_s}{\sqrt{2}} \]

**Calculation:**
\[ V_{rms} = \frac{70 \, \text{V}}{\sqrt{2}} \approx \frac{70}{1.414} \approx 49.5 \, \text{V} \]

### Step 4: Round to the nearest whole number
The problem asks for the nearest whole number of the RMS voltage.

**Final Result:**
\[ V_{rms} \approx 50 \, \text{V} \]

### Conclusion
The nearest RMS secondary voltage is **50 V**.

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