Bridge rectifier feeding R and dc source

When a diode bridge feeds an R in series with a dc source $V_{dc}$, the diodes conduct only while $|v_s| > V_{dc}$. Current flows from angle $\alpha$ to $\pi-\alpha$ in each half cycle, where $V_m\sin\alpha = V_{dc}$.

          io    R = 2 ohm
 P o--->--/\/\/--+
 (bridge         |+
  output)       === Vdc = 80 V
                 |-
 N o-------------+
 Diode bridge fed from 120 V rms, 50 Hz

Step 1: Conduction angles

$$\begin{aligned} V_m &= \sqrt2 \times 120 = 169.71\ \text{V} \\ \alpha &= \sin^{-1}\frac{V_{dc}}{V_m} = \sin^{-1}\frac{80}{169.71} = 28.13^\circ = 0.4909\ \text{rad} \end{aligned}$$

Current flows from $28.13^\circ$ to $151.87^\circ$, i.e. for $123.75^\circ$ (2.1598 rad) in every half cycle:

$$i_o(\omega t) = \frac{V_m\sin\omega t - V_{dc}}{R}, \quad \alpha < \omega t < \pi-\alpha$$

Step 2: Average current

$$\begin{aligned} I_o &= \frac{1}{\pi}\int_{\alpha}^{\pi-\alpha}\frac{V_m\sin\omega t - V_{dc}}{R}d(\omega t) = \frac{1}{\pi R}\left[2V_m\cos\alpha - V_{dc}(\pi-2\alpha)\right] \\ &= \frac{1}{2\pi}\left[2(169.71)(0.8819) - 80(2.1598)\right] = \frac{299.33 - 172.79}{6.2832} = 20.14\ \text{A} \end{aligned}$$

Step 3: RMS current

$$I_{rms}^2 = \frac{1}{\pi R^2}\left[\left(\frac{V_m^2}{2}+V_{dc}^2\right)(\pi-2\alpha) + \frac{V_m^2}{2}\sin2\alpha - 4V_mV_{dc}\cos\alpha\right]$$

Substituting ($V_m^2/2 = 14400$, $\sin2\alpha = 0.8315$):

$$\begin{aligned} I_{rms}^2 &= \frac{1}{4\pi}\left[(14400+6400)(2.1598) + 14400(0.8315) - 4(169.71)(80)(0.8819)\right] \\ &= \frac{44924.4 + 11973.3 - 47893.2}{12.566} = 716.6 \\ I_{rms} &= 26.77\ \text{A} \end{aligned}$$

Step 4: Powers

$$\begin{aligned} P_{dc} &= V_{dc}\,I_o = 80 \times 20.14 = 1611.2\ \text{W} \\ P_R &= I_{rms}^2R = 26.77^2 \times 2 = 1433.1\ \text{W} \end{aligned}$$

The ac source supplies $1611.2 + 1433.1 = 3044.3$ W in total.

Answer: Power absorbed by the dc source = 1611 W; power absorbed by R = 1433 W.