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Resonance is a fascinating and important phenomenon in electrical engineering, especially in alternating current (AC) circuits. It occurs when the inductive and capacitive effects in a circuit balance each other perfectly at a particular frequency, called the resonant frequency. At this frequency, the circuit exhibits unique behavior such as maximum or minimum impedance, which greatly affects current and voltage distribution.
Understanding resonance helps engineers design circuits for radio tuning, filters, power factor correction, and many other applications. In this section, we will explore resonance from the ground up, starting with the fundamental concepts of inductance, capacitance, and reactance, and then move on to series and parallel resonance, their parameters, and practical uses.
Before diving into resonance, it is essential to understand the basic elements that cause it: inductors and capacitors, and how they respond to AC signals.
An inductor is a coil of wire that stores energy in a magnetic field when current flows through it. The property of an inductor that resists changes in current is called inductance, measured in henrys (H).
When AC flows through an inductor, it opposes changes in current by producing a voltage that leads the current by 90°. This opposition is called inductive reactance (\(X_L\)) and depends on frequency:
A capacitor stores energy in an electric field between two conductive plates separated by an insulator. Its ability to store charge is called capacitance, measured in farads (F).
In AC circuits, capacitors oppose changes in voltage by allowing current to lead voltage by 90°. This opposition is called capacitive reactance (\(X_C\)) and decreases as frequency increases:
The key to resonance lies in how \(X_L\) and \(X_C\) vary with frequency. As frequency increases, inductive reactance \(X_L\) increases linearly, while capacitive reactance \(X_C\) decreases inversely.
This opposite frequency dependence is what allows \(X_L\) and \(X_C\) to be equal at a specific frequency, leading to resonance.
Consider a series circuit with a resistor (R), inductor (L), and capacitor (C) connected end-to-end and supplied by an AC source. This is called a series RLC circuit.
Condition for Series Resonance: Resonance occurs when the inductive reactance equals the capacitive reactance:
At this frequency, the inductive and capacitive voltages cancel out, leaving only the resistance to oppose current. Therefore, the circuit's impedance is minimum (equal to R), and the current is maximum.
Key Characteristics of Series Resonance:
Now consider a circuit where the inductor and capacitor are connected in parallel with each other, and this combination is connected in series with a resistor and an AC source. This is called a parallel RLC circuit.
Condition for Parallel Resonance: Resonance occurs when the inductive and capacitive admittances (inverse of reactances) are equal, or equivalently, when the inductive reactance equals the capacitive reactance:
At resonance, the circuit's admittance is minimum, so impedance is maximum, and the current drawn from the source is minimum.
Key Characteristics of Parallel Resonance:
Resonance is not just about the resonant frequency; the sharpness and selectivity of resonance are equally important. Two parameters describe these:
The quality factor \(Q\) measures how sharp or selective the resonance is. A high \(Q\) means the circuit resonates strongly at \(f_0\) and quickly attenuates frequencies away from resonance.
For a series RLC circuit, \(Q\) is defined as:
The bandwidth is the range of frequencies around \(f_0\) where the circuit effectively resonates. It is inversely proportional to \(Q\):
The frequencies \(f_1\) and \(f_2\) are the half-power points where the power drops to half its maximum value (or current drops to \(1/\sqrt{2}\) of maximum).
Step 1: Convert units to standard SI units.
\(L = 10\,\text{mH} = 10 \times 10^{-3} = 0.01\,\text{H}\)
\(C = 100\,\text{nF} = 100 \times 10^{-9} = 1 \times 10^{-7}\,\text{F}\)
Step 2: Use the resonant frequency formula:
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.01 \times 1 \times 10^{-7}}} \]
Step 3: Calculate the denominator:
\(\sqrt{0.01 \times 1 \times 10^{-7}} = \sqrt{1 \times 10^{-9}} = 1 \times 10^{-4.5} = 3.16 \times 10^{-5}\)
Step 4: Calculate \(f_0\):
\[ f_0 = \frac{1}{2\pi \times 3.16 \times 10^{-5}} = \frac{1}{1.986 \times 10^{-4}} \approx 5033\,\text{Hz} \]
Answer: The resonant frequency is approximately 5.03 kHz.
Step 1: Convert units:
\(L = 50\,\text{mH} = 0.05\,\text{H}\)
\(C = 20\,\mu\text{F} = 20 \times 10^{-6} = 2 \times 10^{-5}\,\text{F}\)
Step 2: Calculate resonant frequency \(f_0\):
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.05 \times 2 \times 10^{-5}}} \]
\(\sqrt{0.05 \times 2 \times 10^{-5}} = \sqrt{1 \times 10^{-6}} = 1 \times 10^{-3}\)
\[ f_0 = \frac{1}{2\pi \times 10^{-3}} = \frac{1}{0.006283} \approx 159.15\,\text{Hz} \]
Step 3: Calculate quality factor \(Q\):
\[ Q = \frac{1}{R} \sqrt{\frac{L}{C}} = \frac{1}{10} \sqrt{\frac{0.05}{2 \times 10^{-5}}} \]
\(\frac{0.05}{2 \times 10^{-5}} = 2500\), so \(\sqrt{2500} = 50\)
\[ Q = \frac{1}{10} \times 50 = 5 \]
Step 4: Calculate bandwidth \(BW\):
\[ BW = \frac{f_0}{Q} = \frac{159.15}{5} = 31.83\,\text{Hz} \]
Answer: The bandwidth is approximately 31.8 Hz and the quality factor is 5.
Step 1: Convert units:
\(L = 20\,\text{mH} = 0.02\,\text{H}\)
\(C = 10\,\mu\text{F} = 10 \times 10^{-6} = 1 \times 10^{-5}\,\text{F}\)
Step 2: Use the formula for impedance at resonance in parallel RLC:
Step 3: Calculate \(Z\):
\[ Z = \frac{0.02}{(1 \times 10^{-5}) \times 100} = \frac{0.02}{0.001} = 20\,\Omega \]
Answer: The impedance at resonance is approximately 20 Ω.
Step 1: Calculate the current drawn by the load:
Power \(P = 10,000\,\text{W}\), Voltage \(V = 230\,\text{V}\), Power factor \(pf = 0.7\)
Apparent power \(S = \frac{P}{pf} = \frac{10,000}{0.7} \approx 14,286\,\text{VA}\)
Current \(I = \frac{S}{V} = \frac{14,286}{230} \approx 62.1\,\text{A}\)
Step 2: Calculate reactive power \(Q\):
\[ Q = S \sin \theta = S \sqrt{1 - pf^2} = 14,286 \times \sqrt{1 - 0.7^2} = 14,286 \times 0.714 = 10,204\,\text{VAR} \]
Step 3: Calculate inductive reactance \(X_L\):
\[ X_L = \frac{V}{I \sin \theta} = \frac{230}{62.1 \times 0.714} \approx 5.18\,\Omega \]
Step 4: Calculate inductance \(L\):
\[ L = \frac{X_L}{2\pi f} = \frac{5.18}{2\pi \times 50} = \frac{5.18}{314.16} \approx 0.0165\,\text{H} = 16.5\,\text{mH} \]
Step 5: To correct power factor to unity, add a capacitor with capacitive reactance equal to \(X_L\):
\[ X_C = X_L = 5.18\,\Omega \]
Step 6: Calculate required capacitance \(C\):
\[ C = \frac{1}{2\pi f X_C} = \frac{1}{2\pi \times 50 \times 5.18} = \frac{1}{1626} \approx 6.15 \times 10^{-4}\,\text{F} = 615\,\mu\text{F} \]
Answer: A capacitor of approximately 615 μF is required for power factor correction.
Step 1: Given:
Step 2: Use resonant frequency formula to find \(C\):
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \Rightarrow C = \frac{1}{(2\pi f_0)^2 L} \]
Step 3: Calculate denominator:
\[ (2\pi f_0)^2 = (2 \times 3.1416 \times 1 \times 10^{6})^2 = (6.2832 \times 10^{6})^2 = 3.947 \times 10^{13} \]
Step 4: Calculate \(C\):
\[ C = \frac{1}{3.947 \times 10^{13} \times 1 \times 10^{-6}} = \frac{1}{3.947 \times 10^{7}} = 2.533 \times 10^{-8}\,\text{F} = 25.33\,\text{nF} \]
Step 5: Estimate cost:
Convert capacitance to μF: \(25.33\,\text{nF} = 0.02533\,\mu\text{F}\)
Cost = \(0.02533 \times 500 = Rs.12.67\)
Answer: Required capacitance is approximately 25.3 nF and estimated cost is Rs.13.
When to use: Calculating resonant frequency in any RLC circuit.
When to use: When time is limited and you need to quickly find bandwidth.
When to use: To quickly identify circuit behavior at resonance during problem solving.
When to use: When solving practical problems involving power factor improvement.
When to use: Always, especially during quick exam calculations.
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