A half wave controlled rectifier circuit supplies power to a coil having R=8Ohm, L=50
mH. The input power source is 220V, 50Hz and the firing angle is 60 0 . Derive the
necessary expression and calculate the DC current in the coil.

### Overview:
In a half-wave controlled rectifier, a diode conducts during the positive half cycle of the AC waveform, and the current flows through a resistive-inductive (R-L) load. The firing angle (α) controls when the thyristor (SCR) is triggered, which modifies the output waveform.

#### Given Parameters:
- Resistance, \( R = 8 \, \Omega \)
- Inductance, \( L = 50 \, \text{mH} = 0.05 \, \text{H} \)
- Source voltage, \( V_m = 220 \, \text{V (rms)} \)
- Firing angle, \( \alpha = 60^\circ \)
- Frequency, \( f = 50 \, \text{Hz} \) (hence \( \omega = 2 \pi f = 100 \pi \, \text{rad/s} \))

#### 1. RMS to Peak Voltage:
For an AC source, the peak voltage (\( V_m \)) is related to the RMS voltage (\( V_{\text{rms}} \)) by:
\[
V_m = V_{\text{rms}} \times \sqrt{2}
\]
\[
V_m = 220 \times \sqrt{2} \approx 311 \, \text{V}
\]

### 2. Current Calculation:
To calculate the DC current in the coil, we need to derive the expression for the output current. The general voltage expression for an R-L load is given by:

\[
V(t) = V_m \sin(\omega t)
\]
The equation for the current in an R-L load is governed by the differential equation:

\[
V(t) = R i(t) + L \frac{di(t)}{dt}
\]

Using Laplace transforms or standard methods for solving differential equations for an R-L circuit under half-wave rectification, the solution is known to be:

\[
i(t) = \frac{V_m}{\sqrt{R^2 + (\omega L)^2}} \left[ \sin(\omega t - \theta) - \sin(\alpha - \theta) e^{-\frac{R}{L}(t - \frac{\alpha}{\omega})} \right]
\]

Where:
- \( \theta \) is the impedance angle: 
\[
\theta = \tan^{-1} \left( \frac{\omega L}{R} \right)
\]

Let’s compute \( \theta \):
\[
\theta = \tan^{-1} \left( \frac{\omega L}{R} \right) = \tan^{-1} \left( \frac{100 \pi \times 0.05}{8} \right) \approx \tan^{-1}(1.963) \approx 63.25^\circ
\]

#### 3. Average DC Current Expression:
The DC current can be calculated by taking the average value of the current over one cycle. For a half-wave rectifier, the average current \( I_{DC} \) can be approximated by:

\[
I_{DC} = \frac{V_m}{2 \pi \sqrt{R^2 + (\omega L)^2}} \left(1 + \cos \alpha \right)
\]

Substituting values:
\[
I_{DC} = \frac{311}{2 \pi \sqrt{8^2 + (100 \pi \times 0.05)^2}} \left( 1 + \cos 60^\circ \right)
\]
\[
I_{DC} = \frac{311}{2 \pi \sqrt{64 + (15.71)^2}} \times \left(1 + 0.5 \right)
\]
\[
I_{DC} = \frac{311}{2 \pi \times 16.95} \times 1.5
\]
\[
I_{DC} \approx \frac{311 \times 1.5}{106.46} \approx 4.38 \, \text{A}
\]

### Final Answer:
The DC current in the coil is approximately **4.38 A**.

### Would you like further details on the derivation of the current or any additional clarifications?

### Relative Questions:
1. How does the firing angle impact the DC current in a controlled rectifier?
2. What is the significance of the impedance angle (\( \theta \)) in AC circuits?
3. How would the current change if the firing angle were reduced to \( 45^\circ \)?
4. What effect does increasing inductance have on the performance of an R-L load in a rectifier circuit?
5. Can you explain how a half-wave rectifier differs from a full-wave rectifier in terms of current output?

### Tip:
In circuits with high inductance, the current tends to lag the voltage significantly, and a large firing angle can reduce the average output current significantly.

A half wave controlled rectifier circuit supplies power to a coil having R=8Ohm, L=50mH. The input power source is 220V, 50Hz and the firing angle is 60°. Derive the necessary expression and calculate the DC current in the coil.

Learn how to calculate the DC current in a half-wave controlled rectifier circuit powering an R-L load with R=8Ω, L=50mH, a 220V power source, and a firing angle of 60°. The solution covers AC circuit theory, impedance calculation, and Laplace transform methods for solving R-L circuits.

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