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.