(i) $I_{DC} =\dfrac{1}{\pi}\int_{0}^{\pi}I_{m}\sin \theta d\theta = \dfrac{I_{m}}{\pi}[-\cos \theta]^{\pi}_0 = \dfrac{2I_m}{\pi}$

(ii) Similarly, $V_{DC}=\dfrac{2V_m}{\pi}$

(iii) $I_{rms}^2 = \left [ \dfrac{1}{\pi}\int_{0}^{\pi}I_m^2 \sin^2\theta d\theta \right ]$

$=\dfrac{1}{\pi}.I_m^2\int_{0}^{\pi}\dfrac{1-\cos 2\theta}{2}.d\theta = \dfrac{1}{\pi}I_m^2\left [ \theta-\dfrac{\sin 2\theta}{2} \right ]_{\pi}^0\\ =\dfrac{I_m^2}{2\pi}.[\pi-0] = \dfrac{I_m^2}{2} \Rightarrow I_{rms} = \dfrac{I_m}{\sqrt{2}}$

(iv) $\eta= \dfrac{P_{DC}}{P_{AC}}=\dfrac{I_{DC}^2 \times R_L}{I_{rms}^2 \times (R_L+R_D)} = \dfrac{(2I_m/\pi)^2 \times R_L}{I_{m}^2/2 …