---
title: "Bridge rectifier feeding R and dc source"
subject: "Power Electronics"
chapter: 2
key: "semester_7/power_electronics:bridge-rectifier-dc-source-load"
source: https://www.bctengineer.com/q/semester_7/power_electronics/bridge-rectifier-dc-source-load
site: BCT Engineering
---

# Bridge rectifier feeding R and dc source

Power Electronics · Chapter 2: Single phase AC to DC conversion

- Asked in: 2075 Chaitra, 8 marks
- Video: [Single Phase Full Wave Controlled Rectifier with RLE Load | Power Electronics | Lecture 46](https://www.youtube.com/watch?v=4OqH3R5uNJ8) — Electronics Maddy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/bridge-rectifier-dc-source-load
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-2

## Question

For the full-wave bridge rectifier circuit of Figure below, the ac source is 120 V rms at 50 Hz, R = 2 Ohm and Vdc = 80 V. Determine the power absorbed by the dc voltage source and the power absorbed by the load resistor R. [Figure: single-phase diode bridge (four diodes) from the ac source; output vo drives current io through R in series with a dc source Vdc]

## Short answer

Diodes conduct when |vs|>80 V, α=28.13°. Average current Io=20.14 A, rms current Irms=26.77 A. Power absorbed by the DC source is 1611 W; power absorbed by the resistor is 1433 W.

## Answer

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 $V_m\sin\alpha = V_{dc}$.

```text
          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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. When do the diodes conduct when the load has a dc source Vdc = 80 V?
   A. Whole cycle
   B. Only when |vs| exceeds Vdc, from alpha to pi - alpha
   C. Only when vs is below Vdc
   D. Only at the peak instant
   Answer: B. The dc source keeps the diodes reverse biased until |vs| exceeds Vdc. (Key point: Conduction condition)

2. What is alpha for Vm = 169.71 V and Vdc = 80 V?
   A. alpha = cos^-1(80/169.71) = 61.9°
   B. alpha = sin^-1(80/169.71) = 28.13°
   C. alpha = 45°
   D. alpha = 0°
   Answer: B. Conduction starts when Vm sin alpha = Vdc, giving 28.13°. (Key point: Conduction angle)

3. How is the power absorbed by the dc source found, and what is it?
   A. Vdc x Irms = 2142 W
   B. Vdc x average current = 80 x 20.14 = 1611 W
   C. Irms^2 x Vdc
   D. Vs x Io = 2417 W
   Answer: B. A constant-voltage source absorbs Vdc times the average current. (Key point: Power in dc source)

4. How is the power in R found?
   A. Iavg^2 R
   B. Vdc^2/R
   C. Irms^2 R = 1433 W using the rms of the pulsed current
   D. Vs^2/R
   Answer: C. Resistor loss depends on the rms current, which exceeds the average because the current is pulsed. (Key point: Power in resistor)

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