Understanding Thyristor-Controlled AC Circuits

The question asks us to calculate the power consumed by a load resistor when a 240 V single-phase AC supply feeds it through a thyristor, fired at a specific angle. This setup describes a single-phase half-wave controlled rectifier circuit with a resistive load. A thyristor (SCR) acts like a controlled switch, allowing current to flow only after it is "fired" (triggered) at a certain point in the AC cycle, defined by the firing angle (\(\alpha\)).

Key Parameters for Power Calculation

Let's list the given parameters:

• Supply Voltage (\(V_s\)): 240 V (RMS value of the single-phase AC supply)
• Load Resistor (\(R\)): 100 \(\Omega\)
• Thyristor Firing Angle (\(\alpha\)): 90\(\degree\)

Our goal is to find the power consumed by the load, which can be calculated using the formula \(P = \frac{V_{rms}^2}{R}\), where \(V_{rms}\) is the RMS voltage across the load resistor.

Calculating Peak Supply Voltage

First, we need to convert the RMS supply voltage to its peak value. For a sinusoidal AC supply, the peak voltage (\(V_{peak}\)) is related to the RMS voltage (\(V_s\)) by:

\[ V_{peak} = V_s \times \sqrt{2} \]

Substituting the given RMS supply voltage:

\[ V_{peak} = 240 \text{ V} \times \sqrt{2} \approx 240 \text{ V} \times 1.414 = 339.36 \text{ V} \]

Determining RMS Voltage Across the Load

For a single-phase half-wave controlled rectifier with a resistive load, the RMS voltage across the load (\(V_{rms}\)) is given by the formula:

\[ V_{rms}^2 = \frac{V_{peak}^2}{4\pi} \left( \pi - \alpha + \frac{\sin(2\alpha)}{2} \right) \]

Where \(\alpha\) must be in radians. Given \(\alpha = 90^\degree\), we convert it to radians:

\[ \alpha = 90^\degree = \frac{\pi}{2} \text{ radians} \]

Now, substitute \(\alpha = \frac{\pi}{2}\) into the RMS voltage formula:

\[ V_{rms}^2 = \frac{V_{peak}^2}{4\pi} \left( \pi - \frac{\pi}{2} + \frac{\sin\left(2 \times \frac{\pi}{2}\right)}{2} \right) \]

\[ V_{rms}^2 = \frac{V_{peak}^2}{4\pi} \left( \frac{\pi}{2} + \frac{\sin(\pi)}{2} \right) \]

Since \(\sin(\pi) = 0\):

\[ V_{rms}^2 = \frac{V_{peak}^2}{4\pi} \left( \frac{\pi}{2} + 0 \right) \]

\[ V_{rms}^2 = \frac{V_{peak}^2}{4\pi} \times \frac{\pi}{2} \]

\[ V_{rms}^2 = \frac{V_{peak}^2}{8} \]

Taking the square root of both sides, we get:

\[ V_{rms} = \frac{V_{peak}}{\sqrt{8}} = \frac{V_{peak}}{2\sqrt{2}} \]

We know that \(V_{peak} = V_s \sqrt{2}\). Substitute this into the expression for \(V_{rms}\):

\[ V_{rms} = \frac{V_s \sqrt{2}}{2\sqrt{2}} = \frac{V_s}{2} \]

Now, substitute the given RMS supply voltage \(V_s = 240 \text{ V}\):

\[ V_{rms} = \frac{240 \text{ V}}{2} = 120 \text{ V} \]

Calculating Power Consumed by the Load

Finally, we can calculate the power consumed by the load resistor using the formula \(P = \frac{V_{rms}^2}{R}\):

\[ P = \frac{(120 \text{ V})^2}{100 \text{ } \Omega} \]

\[ P = \frac{14400 \text{ V}^2}{100 \text{ } \Omega} \]

\[ P = 144 \text{ W} \]

Summary of Calculation Steps

The following table summarizes the calculations:

Therefore, the power consumed by the load is 144 W.