(i) RMS output voltage can be obtained from

$$E_{0 \mathrm{rms}}=\left[\frac{2}{T / 2} \int_{0}^{T / 2} E^{2} \mathrm{d} t\right]^{1 / 2}$$

$$\therefore \quad E_{0 \operatorname{rms}}=E_{\mathrm{dc}}-----(1)$$

(ii) The instantaneous output voltage can be expressed in fourier series as

$$e_{0(wt)}=\sum_{n=1,3,5, \ldots}^{\infty} \frac{4 E_{\mathrm{dc}}}{n \pi} \sin n \omega t-----(2)$$

The output voltage waveform contains only the odd harmonic components, i.e. $3,5,7, \ldots$ The even order harmonics are automatically cancelled.

(iii) For $n=1,$ Eq. (2) gives the rms value of the fundamental component

$$E_{1(\operatorname{rms})}=\frac{4 E_{\mathrm{dc}}}{\sqrt{2} \cdot \pi}=0.9 E_{\mathrm{dc}}-----(3)$$

(iv) For RL load, the equation for the instantaneous current $i_{0}$ can be found using the Equation $(2),$ as

$$i_{0(t)}=\sum_{n=1,3,5, \ldots}^{\infty} \frac{4 E_{\mathrm{dc}}}{n \pi \sqrt{R^{2}+(n \omega L)^{2}}} \sin \left(n \omega t-\theta_{n}\right)$$

In this equation, $Z_{n}=\sqrt{R^{2}+(n \omega L)^{2}}$ is the impedance offered by the load to the $n^{\text { th }}$ harmonic component and $\frac{4 E_{\mathrm{dc}}}{n \pi}$ is the peak amplitude of $n^{th}$ harmonic voltage, and

$$\theta_{n}=\tan ^{-1}(n \omega L / R)-----(5)$$