The r.m.s value of alternating e.m.f. is

Options

A. twice peak value
B. `sqrt(2)` times greater than peak value
C. equal to peak value
D. less than peak value

Answer

To find the r.m.s (root mean square) value of an alternating e.m.f., we can follow these steps:

### Step 1: Understand the Definition of r.m.s Value
The r.m.s value of any periodic function is defined as:
\[ V_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T V(t)^2 \, dt} \]
where \( T \) is the time period of the function and \( V(t) \) is the instantaneous value of the voltage.

### Step 2: Define the Alternating Voltage
For an alternating e.m.f., we can express it as:
\[ V(t) = V_0 \sin(\omega t) \]
where \( V_0 \) is the peak voltage and \( \omega \) is the angular frequency.

### Step 3: Substitute into the r.m.s Formula
Substituting \( V(t) \) into the r.m.s formula gives:
\[ V_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T (V_0 \sin(\omega t))^2 \, dt} \]
This simplifies to:
\[ V_{\text{rms}} = \sqrt{\frac{V_0^2}{T} \int_0^T \sin^2(\omega t) \, dt} \]

### Step 4: Evaluate the Integral
The integral \( \int_0^T \sin^2(\omega t) \, dt \) can be evaluated using the identity:
\[ \sin^2(x) = \frac{1 - \cos(2x)}{2} \]
Thus,
\[ \int_0^T \sin^2(\omega t) \, dt = \int_0^T \frac{1 - \cos(2\omega t)}{2} \, dt \]
This separates into two integrals:
\[ = \frac{1}{2} \int_0^T 1 \, dt - \frac{1}{2} \int_0^T \cos(2\omega t) \, dt \]
The first integral evaluates to \( \frac{T}{2} \) and the second integral evaluates to zero (since \( \cos(2\omega t) \) completes a full cycle over the interval \( [0, T] \)):
\[ \int_0^T \cos(2\omega t) \, dt = 0 \]
Therefore,
\[ \int_0^T \sin^2(\omega t) \, dt = \frac{T}{2} \]

### Step 5: Substitute Back into the r.m.s Formula
Substituting this result back into the r.m.s formula:
\[ V_{\text{rms}} = \sqrt{\frac{V_0^2}{T} \cdot \frac{T}{2}} = \sqrt{\frac{V_0^2}{2}} = \frac{V_0}{\sqrt{2}} \]

### Step 6: Conclusion
Thus, the r.m.s value of the alternating e.m.f. is:
\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \]