The effective value of the current for the Sawtooth waveform as shown is: 
$\frac{I_m}{\sqrt{3}}$
To find the effective (RMS) value of a sawtooth waveform, we can use the formula for the RMS value of such non-sinusoidal waveforms. The RMS value is calculated as:
\(I_{\text{RMS}} = \sqrt{\frac{1}{T} \int_0^T i^2(t) \, dt}\)
Given the piecewise linear nature of the sawtooth waveform shown in the figure, the instantaneous current \(i(t)\) can be expressed as a linear function of time.
The equation for the sawtooth waveform between 0 and T is:
\(i(t) = \frac{I_m}{T} \cdot t\)
Substitute this into the RMS formula:
\(I_{\text{RMS}} = \sqrt{\frac{1}{T} \int_0^T \left(\frac{I_m}{T} \cdot t\right)^2 \, dt}\)
\(= \sqrt{\frac{1}{T} \times \frac{I_m^2}{T^2} \int_0^T t^2 \, dt}\)
\(= \sqrt{\frac{I_m^2}{T^3} \left[\frac{t^3}{3}\Bigg|_0^T\right]}\)
\(= \sqrt{\frac{I_m^2}{T^3} \cdot \frac{T^3}{3}}\)
\(= \sqrt{\frac{I_m^2}{3}}\)
\(= \frac{I_m}{\sqrt{3}}\)
Thus, the effective value of the current for the sawtooth waveform is \(\frac{I_m}{\sqrt{3}}\), which matches the given correct answer.
Therefore, the correct option is:
- \(\frac{I_m}{\sqrt{3}}\)
Important Questions from Fourier Series
The Fourier series representation of a square wave is shown in the figure below. The fluctuations seen near $x = \pm 1$ are named after which one of the following scientists?
The Fourier series expansion of the function $f(x) = |\cos(x)|$ in the interval $(-\pi, \pi)$ is $\alpha + \beta \left[ \frac{1}{3}\cos(2x) - \frac{1}{15}\cos(4x) + \dots \right]$.
The values of $\alpha$ and $\beta$ respectively are ______.
Periodic function $f(x)$ is given below. $$f(x) = \begin{cases} -1, & \text{when } -\pi < x < 0 \\ 1, & \text{when } 0 < x < \pi \end{cases} ; f(x + 2\pi) = f(x)$$ The CORRECT option representing the Fourier series expansion of $f(x)$ is:
A time-limited waveform $g(x)$ is specified as follows: $$g(x) = \begin{cases} -k, & -\pi < x \le 0 \\ +k, & 0 < x \le \pi \\ 0, & \text{otherwise} \end{cases}$$ A new waveform $f(x)$ is constructed from $g(x)$ as follows: $$f(x) = \sum_{m=-\infty}^{\infty} g(x + 2\pi m), \quad \text{for all } x \in \mathbb{R}$$ The sum of the coefficients of the third harmonics of the sine and cosine terms in the trigonometric Fourier series expansion of $f(x)$ is $\frac{2}{3\pi}$.
What is the value of $k$?
For fourier series of wave shown in Figure below,
Select correct expression for f(t)
