Power Dissipation in Series LCR Circuit at Resonant Condition

In a series LCR circuit, understanding the behavior at resonant condition is crucial for calculating various electrical parameters, including power dissipation. At resonance, the inductive reactance (\(X_L\)) becomes equal to the capacitive reactance (\(X_C\)). This equality leads to the total impedance of the circuit being purely resistive.

Understanding Resonant Condition

When a series LCR circuit is at resonance, two key conditions are met:

• The inductive reactance, \(X_L = \omega L\), is equal to the capacitive reactance, \(X_C = \frac{1}{\omega C}\).
• As a result, the phase difference between the voltage and current in the circuit becomes zero.

The total impedance (\(Z\)) of the series LCR circuit is given by the formula:

\(Z = \sqrt{R^2 + (X_L - X_C)^2}\)

At resonance, since \(X_L = X_C\), the term \((X_L - X_C)\) becomes zero. Therefore, the impedance simplifies to:

\(Z = \sqrt{R^2 + 0^2} = R\)

This means that at resonance, the circuit behaves purely resistively, and the current in the circuit is maximum for a given voltage.

Given Parameters for the LCR Circuit

Let's list the given parameters from the problem statement:

• Peak value of sinusoidal voltage, \(V_p = 250 \, \text{V}\)
• Resistance, \(R = 8 \, \Omega\)
• Inductance, \(L = 24 \, \text{mH} = 24 \times 10^{-3} \, \text{H}\)
• Capacitance, \(C = 60 \, \mu\text{F} = 60 \times 10^{-6} \, \text{F}\)

Calculating RMS Voltage

The power dissipation is usually calculated using RMS (Root Mean Square) values of voltage and current. The relationship between peak voltage (\(V_p\)) and RMS voltage (\(V_{rms}\)) for a sinusoidal waveform is:

\(V_{rms} = \frac{V_p}{\sqrt{2}}\)

Substituting the given peak voltage:

\(V_{rms} = \frac{250}{\sqrt{2}} \, \text{V}\)

Determining Power Dissipation at Resonant Condition

The power dissipated in an AC circuit is given by the formula:

\(P = V_{rms} I_{rms} \cos(\phi)\)

Where \(\cos(\phi)\) is the power factor. At resonance, as discussed earlier, the circuit is purely resistive, which means the phase angle \(\phi = 0^\circ\). Therefore, \(\cos(\phi) = \cos(0^\circ) = 1\).

So, the power dissipated at resonant condition simplifies to:

\(P = V_{rms} I_{rms}\)

Also, at resonance, the impedance \(Z = R\). According to Ohm's law, \(I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{rms}}{R}\).

Substituting \(I_{rms}\) into the power formula:

\(P = V_{rms} \left(\frac{V_{rms}}{R}\right)\)

\(P = \frac{V_{rms}^2}{R}\)

Now, substitute the calculated \(V_{rms}\) and the given resistance \(R\):

\(P = \frac{\left(\frac{250}{\sqrt{2}}\right)^2}{8}\)

\(P = \frac{\frac{250^2}{2}}{8}\)

\(P = \frac{62500}{2 \times 8}\)

\(P = \frac{62500}{16}\)

\(P = 3906.25 \, \text{W}\)

Value of 'X' in Kilowatts

The problem asks for the power dissipated in kilowatts, denoted as 'X' kW. To convert Watts to Kilowatts, we divide by 1000:

\(P_{kW} = \frac{3906.25}{1000} \, \text{kW}\)

\(P_{kW} = 3.90625 \, \text{kW}\)

So, the value of 'X' is 3.90625.

The problem asks for the value of 'X' to the nearest integer. Rounding 3.90625 to the nearest integer gives 4.

Thus, the value of 'X' is 4.