Question

An ac voltage \(V = 280 \sin (100\pi t)\) volt is connected across a series LCR circuit in which \(R = 400 \, \Omega\), \(L = \frac{5}{\pi} \, \text{H}\) and \(C = \frac{50}{\pi} \, \mu\text{F}\). Taking \(\sqrt{2} = 1.4\), calculate :
(I) impedance of the circuit.
(II) rms value of current that flows in the circuit.
(III) power factor of the circuit.

Hint:

Whenever you encounter values like 300 and 400 for orthogonal components, remember the standard Pythagorean triplet (3, 4, 5). Here, \(\sqrt{400^2 + 300^2}\) instantly reveals itself to be 500, reducing calculation errors significantly during tests!

Answer 1

Concept: For a series LCR circuit driven by an alternating electromotive force, the fundamental quantities are evaluated using standard AC analysis.
• Standard AC Voltage Equation: \(V(t) = V_0 \sin(\omega t)\), where \(V_0\) is the peak voltage amplitude and \(\omega\) is the angular frequency in radians per second.
• Inductive Reactance: \(X_L = \omega L\)
• Capacitive Reactance: \(X_C = \frac{1}{\omega C}\)
• Total Impedance (\(Z\)): Combining the resistive, inductive, and capacitive parameters in a vector space gives: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
• RMS Values: The root-mean-square values for voltage and current represent their effective DC equivalents and are linked by: \[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}}, \quad I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} \]
• Power Factor: The power factor is defined as the cosine of the phase angle (\(\phi\)) between the current and voltage, which is equivalent to the ratio of pure resistance to total impedance: \[ \text{Power Factor} = \cos\phi = \frac{R}{Z} \]

Step 1: Extract and identify all given numerical values from the source parameters.
Comparing the provided instantaneous voltage expression \(V = 280 \sin (100\pi t)\) with the standard general form \(V = V_0 \sin(\omega t)\), we can map out the values directly:
• Peak value of alternating voltage, \(V_0 = 280 \, \text{V}\)
• Angular frequency of the source, \(\omega = 100\pi \, \text{rad/s}\)
• Pure resistance of the circuit, \(R = 400 \, \Omega\)
• Inductance of the inductor, \(L = \frac{5}{\pi} \, \text{H}\)
• Capacitance of the capacitor, \(C = \frac{50}{\pi} \, \mu\text{F} = \frac{50}{\pi} \times 10^{-6} \, \text{F}\)
• Approximation constant given: \(\sqrt{2} = 1.4\)

Step 2: Calculate the inductive reactance (\(X_L\)) and capacitive reactance (\(X_C\)).
Let us calculate the individual reactances explicitly: \[ X_L = \omega L = (100\pi) \times \left(\frac{5}{\pi}\right) \] Canceling out the transcendental constant \(\pi\) from both numerator and denominator: \[ X_L = 100 \times 5 = 500 \, \Omega \] Next, computing the capacitive reactance: \[ X_C = \frac{1}{\omega C} = \frac{1}{(100\pi) \times \left(\frac{50}{\pi} \times 10^{-6}\right)} \] Canceling out the \(\pi\) terms in the denominator: \[ X_C = \frac{1}{100 \times 50 \times 10^{-6}} = \frac{1}{5000 \times 10^{-6}} \] Expressing the denominator in simplified power-of-ten scientific notation: \[ 5000 \times 10^{-6} = 5 \times 10^3 \times 10^{-6} = 5 \times 10^{-3} \] Substituting this back into the fraction: \[ X_C = \frac{1}{5 \times 10^{-3}} = \frac{10^3}{5} = \frac{1000}{5} = 200 \, \Omega \] Thus, we have successfully evaluated: \[ X_L = 500 \, \Omega \quad \text{and} \quad X_C = 200 \, \Omega \]

Step 3: Calculate (I) the total impedance of the circuit (\(Z\)).
Using the standard impedance formula for a series combination circuit: \[ Z = \sqrt{R^2 + (X_L - X_C)^2} \] Substitute our values (\(R = 400 \, \Omega\), \(X_L = 500 \, \Omega\), \(X_C = 200 \, \Omega\)) into this expression: \[ Z = \sqrt{(400)^2 + (500 - 200)^2} \] Subtracting the reactances inside the parenthesis: \[ 500 - 200 = 300 \, \Omega \] Now, evaluating the squares of both individual terms: \[ (400)^2 = 160000 \] \[ (300)^2 = 90000 \] Summing the values inside the radical sign: \[ Z = \sqrt{160000 + 90000} = \sqrt{250000} \] Taking the square root explicitly: \[ Z = 500 \, \Omega \] Answer for (I): The impedance of the circuit is \(500 \, \Omega\).

Step 4: Calculate (II) the rms value of current that flows in the circuit (\(I_{\text{rms}}\)).
First, we must calculate the root-mean-square value of the alternating voltage source, \(V_{\text{rms}}\): \[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \] Substitute the peak voltage \(V_0 = 280 \, \text{V}\) and the specified value \(\sqrt{2} = 1.4\): \[ V_{\text{rms}} = \frac{280}{1.4} = \frac{2800}{14} = 200 \, \text{V} \] Now, using Ohm's law for AC parameters to evaluate the effective alternating current: \[ I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} \] Substitute \(V_{\text{rms}} = 200 \, \text{V}\) and our calculated impedance \(Z = 500 \, \Omega\): \[ I_{\text{rms}} = \frac{200}{500} = \frac{2}{5} = 0.4 \, \text{A} \] Answer for (II): The rms value of current flowing through the circuit is \(0.4 \, \text{A}\).

Step 5: Calculate (III) the power factor of the circuit.
The formula for the dimensionless power factor of a series network is defined as: \[ \text{Power Factor} = \cos\phi = \frac{R}{Z} \] Substitute the given resistance \(R = 400 \, \Omega\) and the computed impedance total \(Z = 500 \, \Omega\): \[ \cos\phi = \frac{400}{500} = \frac{4}{5} = 0.8 \] Answer for (III): The power factor of the circuit is \(0.8\).