In an ideal transformer the number turns of primary and secondary coil is given as 100 and 200 respectively. If the peak value of the primary voltage is 50V, then the r.m.s value of secondary voltage is nearest to

Options

A. 100V
B. 200V
C. 60V
D. 70V

Answer

To solve the problem, we need to find the r.m.s value of the secondary voltage (Vs_rms) in an ideal transformer given the number of turns in the primary (Np), the number of turns in the secondary (Ns), and the peak value of the primary voltage (Vp_peak).

### Step-by-step Solution:

1. **Identify the Given Values:**
- Number of turns in the primary coil (Np) = 100
- Number of turns in the secondary coil (Ns) = 200
- Peak value of the primary voltage (Vp_peak) = 50 V

2. **Use the Transformer Equation:**
In an ideal transformer, the relationship between the voltages and the number of turns is given by:
\[ \frac{Vp}{Vs} = \frac{Np}{Ns} \]
Rearranging this gives:
\[ Vs = Vp \cdot \frac{Ns}{Np} \]

3. **Calculate the Peak Value of the Secondary Voltage (Vs_peak):**
Substitute the known values into the equation:
\[ Vs_peak = 50 \cdot \frac{200}{100} = 50 \cdot 2 = 100 \text{ V} \]

4. **Convert Peak Voltage to R.M.S Voltage:**
The relationship between the peak voltage (V_peak) and the r.m.s voltage (V_rms) for an AC voltage is given by:
\[ V_rms = \frac{V_peak}{\sqrt{2}} \]
Therefore, for the secondary voltage:
\[ Vs_rms = \frac{Vs_peak}{\sqrt{2}} = \frac{100}{\sqrt{2}} \approx \frac{100}{1.414} \approx 70.7 \text{ V} \]

5. **Round to the Nearest Value:**
The r.m.s value of the secondary voltage is approximately 70.7 V, which we can round to the nearest whole number:
\[ Vs_rms \approx 71 \text{ V} \]

### Final Answer:
The r.m.s value of the secondary voltage is nearest to **71 V**.

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