🔍
A 1-phase Half controlled rectifier supplied with voltage v=300sin 314t, and load resistance is 100Ω. Find Average output voltage Load current (For α=60° and α=100°).

2 Answers

 
Best answer
To find the average output voltage and load current of a 1-phase half-controlled rectifier, we can use the following formulas for a given firing angle \(\alpha\).

### Given:
- Input voltage \( v(t) = 300 \sin(314 t) \)
- Load resistance \( R = 100 \, \Omega \)
- Firing angles \(\alpha = 60^\circ\) and \(\alpha = 100^\circ\)

### 1. **Average Output Voltage \(V_{dc}\)**

The average output voltage for a 1-phase half-controlled rectifier is given by:

\[ V_{dc} = \frac{V_m}{\pi} \left( \cos \alpha + \frac{\pi - \alpha}{2} \sin \alpha \right) \]

where \( V_m \) is the maximum value of the input voltage.

For the given input voltage \( v(t) = 300 \sin(314 t) \), the peak voltage \( V_m \) is 300 V.

#### For \(\alpha = 60^\circ\):

1. Convert \(\alpha\) to radians: \(\alpha = 60^\circ = \frac{\pi}{3} \text{ radians}\).
2. Substitute \( \alpha = \frac{\pi}{3} \) into the formula:

\[ V_{dc} = \frac{300}{\pi} \left( \cos \frac{\pi}{3} + \frac{\pi - \frac{\pi}{3}}{2} \sin \frac{\pi}{3} \right) \]

\[ V_{dc} = \frac{300}{\pi} \left( \frac{1}{2} + \frac{2\pi/3}{2} \cdot \frac{\sqrt{3}}{2} \right) \]

\[ V_{dc} = \frac{300}{\pi} \left( \frac{1}{2} + \frac{\pi \sqrt{3}}{6} \right) \]

\[ V_{dc} = \frac{300}{\pi} \left( \frac{1}{2} + \frac{0.912}{6} \right) \]

\[ V_{dc} \approx \frac{300}{\pi} \times 0.695 \]

\[ V_{dc} \approx 66.4 \text{ V} \]

#### For \(\alpha = 100^\circ\):

1. Convert \(\alpha\) to radians: \(\alpha = 100^\circ = \frac{100\pi}{180} = \frac{5\pi}{9} \text{ radians}\).
2. Substitute \( \alpha = \frac{5\pi}{9} \) into the formula:

\[ V_{dc} = \frac{300}{\pi} \left( \cos \frac{5\pi}{9} + \frac{\pi - \frac{5\pi}{9}}{2} \sin \frac{5\pi}{9} \right) \]

\[ V_{dc} = \frac{300}{\pi} \left( \cos \frac{5\pi}{9} + \frac{4\pi/9}{2} \sin \frac{5\pi}{9} \right) \]

\[ V_{dc} = \frac{300}{\pi} \left( -0.766 + \frac{2\pi/9 \cdot 0.766}{2} \right) \]

\[ V_{dc} \approx \frac{300}{\pi} \left( -0.766 + 0.135 \right) \]

\[ V_{dc} \approx \frac{300}{\pi} \times -0.631 \]

\[ V_{dc} \approx -60.2 \text{ V} \]

### 2. **Load Current \(I_{dc}\)**

The load current is given by:

\[ I_{dc} = \frac{V_{dc}}{R} \]

#### For \(\alpha = 60^\circ\):

\[ I_{dc} = \frac{66.4}{100} \]

\[ I_{dc} \approx 0.664 \text{ A} \]

#### For \(\alpha = 100^\circ\):

\[ I_{dc} = \frac{-60.2}{100} \]

\[ I_{dc} \approx -0.602 \text{ A} \]

### Summary:

- **For \(\alpha = 60^\circ\)**:
  - Average output voltage \(V_{dc} \approx 66.4 \text{ V}\)
  - Load current \(I_{dc} \approx 0.664 \text{ A}\)

- **For \(\alpha = 100^\circ\)**:
  - Average output voltage \(V_{dc} \approx -60.2 \text{ V}\)
  - Load current \(I_{dc} \approx -0.602 \text{ A}\)
0 like 0 dislike
To analyze a 1-phase half-controlled rectifier circuit with a resistive load, you need to determine the average output voltage and load current for different firing angles (\(\alpha\)).

### Given:
- Supply Voltage: \( V_s = 300 \sin(314t) \)
- Load Resistance: \( R = 100 \, \Omega \)
- Firing Angles: \( \alpha = 60^\circ \) and \( \alpha = 100^\circ \)

### Analysis of Half-Controlled Rectifier

A half-controlled rectifier circuit consists of a single SCR (Silicon Controlled Rectifier) and a diode. The output voltage and current depend on the firing angle \(\alpha\).

**1. Peak Voltage Calculation:**

The given supply voltage is \( V_s = 300 \sin(314t) \), where the angular frequency \( \omega \) is \( 314 \) rad/s.

The peak voltage \( V_m \) is given by:
\[ V_m = 300 \, \text{V} \]

**2. Average Output Voltage Calculation:**

The average output voltage \( V_{o_{avg}} \) for a half-controlled rectifier is given by:
\[ V_{o_{avg}} = \frac{V_m}{\pi} \left( \pi - \alpha \right) \]

where \(\alpha\) is the firing angle in radians.

Convert \(\alpha\) from degrees to radians:
\[ \alpha_{rad} = \frac{\alpha \times \pi}{180} \]

**For \(\alpha = 60^\circ\):**
\[ \alpha_{rad} = \frac{60 \times \pi}{180} = \frac{\pi}{3} \]

Substitute \(\alpha_{rad}\) into the formula:
\[ V_{o_{avg}} = \frac{300}{\pi} \left( \pi - \frac{\pi}{3} \right) \]
\[ V_{o_{avg}} = \frac{300}{\pi} \left( \frac{2\pi}{3} \right) \]
\[ V_{o_{avg}} = 300 \times \frac{2}{3} \]
\[ V_{o_{avg}} = 200 \, \text{V} \]

**For \(\alpha = 100^\circ\):**
\[ \alpha_{rad} = \frac{100 \times \pi}{180} = \frac{5\pi}{9} \]

Substitute \(\alpha_{rad}\) into the formula:
\[ V_{o_{avg}} = \frac{300}{\pi} \left( \pi - \frac{5\pi}{9} \right) \]
\[ V_{o_{avg}} = \frac{300}{\pi} \left( \frac{4\pi}{9} \right) \]
\[ V_{o_{avg}} = 300 \times \frac{4}{9} \]
\[ V_{o_{avg}} = 133.33 \, \text{V} \]

**3. Load Current Calculation:**

The load current \( I_{o} \) is given by:
\[ I_{o} = \frac{V_{o_{avg}}}{R} \]

**For \(\alpha = 60^\circ\):**
\[ I_{o} = \frac{200}{100} \]
\[ I_{o} = 2 \, \text{A} \]

**For \(\alpha = 100^\circ\):**
\[ I_{o} = \frac{133.33}{100} \]
\[ I_{o} = 1.333 \, \text{A} \]

### Summary

- **For \(\alpha = 60^\circ\):**
  - Average Output Voltage: \( 200 \, \text{V} \)
  - Load Current: \( 2 \, \text{A} \)

- **For \(\alpha = 100^\circ\):**
  - Average Output Voltage: \( 133.33 \, \text{V} \)
  - Load Current: \( 1.333 \, \text{A} \)
0 like 0 dislike

Related questions

A 20 pole, 693V, 50 Hz, 3 phase, delta connected synchronous motor is operating at no load with normal excitation. It has armature resistance per phase of zero and synchronous reactance of 10 ohm. If ... iii) armature current /phase, iv) power drawn by the motor and v) power developed by armature.
Answer : To solve the problem, let's break it down step by step: ### Given Data - **Number of poles (P)**: 20 - **Voltage (V)**: 693V (line-to-line) - **Frequency (f)**: 50 Hz - ** ... **Power drawn by the motor**: Approximately 4,188 W - **Power developed by the armature**: Approximately 407.6 W...

Show More

Define voltage regulation of alternator. A 400V, 10 KVA, 3 phase star connected alternator has resistance per phase of 1.0 ohm. Open circuit voltage per phase of 90V is obtained for field current of 1.0 ... per phase and iv) Regulation while supplying a load current of 15A at 0.8 power factor lag.
Answer : **Voltage Regulation of Alternator:** Voltage regulation of an alternator refers to the change in terminal voltage when the alternator shifts from no-load to full-load conditions while the field current remains constant. It' ... (\( E_{oc} \))**: 90 V 4. **Voltage Regulation**: Approximately 264.9%...

Show More

Explain with neat circuit diagram and input output waveforms, single phase half wave converter with R-L load. Give significance of freewheeling Diode.
Answer : To delve into the single-phase half-wave converter with an R-L load, let's start with the basic operation and then discuss the significance of a freewheeling diode. ### Circuit Diagram 1. * ... kickback, ensuring smoother operation. If you need more details or specific waveforms, feel free to ask!...

Show More

Rectifier: Line commutated 1 phase and 3 phase half (semi) & fully controlled bridge converters
Answer : A **rectifier** is an electrical device that converts alternating current (AC) into direct current (DC). Rectifiers are essential components in various power supply systems, such as in motor ... level of control required over the output voltage and the ripple tolerance for the specific application....

Show More

A balanced delta connected load is connected to a 415 V, 50 Hz supply. If the per phase impedance of the load is (8 + j12) ohm, then calculate the phase current of the load. Also, find out the line current and the power consumed by each phase.
Answer : To solve this problem, we need to follow these steps: 1. **Calculate the Phase Voltage:** For a balanced delta-connected load, the line voltage \( V_L \) is related to the phase voltage \( V_{ph} \) by the ... \( I_L \): 28.8 A - **Power Consumed by Each Phase** \( P_{ph} \): 2215.2 W...

Show More
Welcome to Electrical Engineering, where you can ask questions and receive answers from other members of the community.