AC Circuit Analysis PDF

Document Details

CalmingBaltimore2166

Uploaded by CalmingBaltimore2166

Parul University

Tags

AC circuits Alternating current Electrical engineering Circuit analysis

Summary

This document provides a detailed explanation of AC circuit analysis focusing on pure resistance, inductance and capacitance. It includes diagrams, equations, and calculations for instantaneous and average power. Ideal for undergraduate-level electrical engineering students.

Full Transcript

# AC circuit with a pure resistance - **Diagram:** A circuit diagram shows an AC circuit with a pure resistance R. An AC voltage source is connected to the resistor. - **Equation for the voltage:** The alternating voltage v is given by: $v = V_m sin(ωt)$ - **Equation for the current:** The...

# AC circuit with a pure resistance - **Diagram:** A circuit diagram shows an AC circuit with a pure resistance R. An AC voltage source is connected to the resistor. - **Equation for the voltage:** The alternating voltage v is given by: $v = V_m sin(ωt)$ - **Equation for the current:** The current flowing in the circuit is i. The voltage across the resistor is given as VR, which is the same as V. Using Ohm's law, the following relations can be written: $i = \frac{v}{R} = \frac{V_m}{R} sin(ωt)$ - **Equation for the current:** The current can be written as: $i = I_m sin(ωt)$ - **Relationship between voltage and current:** The formulas for voltage and current show that in a pure resistive circuit, the voltage and current are in phase. - **Diagram:** Sine waves of voltage and current are in phase. The voltage and current waveforms and phasors can be drawn as below: # Instantaneous power The instantaneous power in the circuit can be derived as follows: $p = vi$ $p = (V_m sin(ωt))(I_m sin(ωt))$ $p = V_m I_m sin^2(ωt)$ <start_of_image> $p = \frac{V_m I_m}{2}(1 - cos(2ωt))$ $p = \frac{V_m I_m}{2} - \frac{V_m I_m}{2} cos(2ωt)$ - **Power terms:** The instantaneous power consists of two terms: * The first is the **constant power term** * The second term is the **fluctuating power term**. # Average power The average power over one cycle can be found as follows: $P = \frac{1}{2π} \int_{0}^{2π} [\frac{V_m I_m}{2} - \frac{V_m I_m}{2} cos(2ωt)]dωt$ $P = \frac{V_m I_m}{2} - \frac{V_m I_m}{2π}[\frac{sin(2ωt)}{2}]_0^{2π}$ $P = \frac{V_m I_m}{2} = \frac{V_m I_m}{√2 √2}$ $P = V_{rms} I_{rms}$ # AC circuit with a pure inductance - **Diagram:** A circuit diagram shows a pure inductance L. An AC voltage source is connected to the inductor. - **Equation for the voltage:** The alternating voltage v is given by: $v = V_m sin(ωt)$ - **Equation for the current:** The current flowing in the circuit is i. The voltage across the inductor is given as VL which is the same as V. The current through the inductor can be found as follows: $v = L\frac{di}{dt} = Ldi$ $V_m sin(ωt) = L\frac{di}{dt}$ $di = \frac{V_m}{L} sin(ωt)dt$ $i = \frac{V_m}{L} \int sin(ωt)dt$ $i = \frac{V_m}{ωL}(-cos(ωt))$ $i = \frac{V_m}{ωL} sin(ωt - \frac{π}{2})$ $i = I_m sin(ωt - \frac{π}{2})$ - **Relationship between voltage and current:** From the formulas for voltage and current, it can be seen that in a pure inductive circuit, the current lags behind the voltage by 90°. - **Diagram:** Sine wave of voltage leads the sine wave of current by 90 degrees. # Inductive reactance The inductive reactance XL is given as: $X_L = ωL = 2πf_L$ # Instantaneous power The instantaneous power in the circuit can be derived as follows: $p = vi$ $p = (V_m sin(ωt))(I_m sin(ωt - \frac{π}{2}))$ $p = -V_m I_m sin(ωt)cos(ωt)$ $p = -\frac{V_m I_m}{2} sin(2ωt)$ - **Nature of power:** The instantaneous power is fluctuating in nature. # Average power The average power over one cycle can be found as follows: $P = \frac{1}{2π} \int_{0}^{2π} [-\frac{V_m I_m}{2} sin(2ωt)]dωt$ $P = 0$ - **Power consumed:** The average power in a pure inductive circuit is zero. In other words, the power consumed by a pure inductance is zero. # AC circuit with a pure capacitance - **Diagram:** A circuit diagram shows a pure capacitance C. An AC voltage source is connected to the capacitor. - **Equation for the voltage:** The alternating voltage v is given by: $v = V_m sin(ωt)$ - **Equation for the current:** The current flowing in the circuit is i. The voltage across the capacitor is given as VC, which is the same as V. The current through the capacitor can be found as follows: $q = CV$ $q = CV_m sin(ωt)$ $i = \frac{dq}{dt}$ $i = CV_m ωcos(ωt)$ $i = ωCV_m sin(ωt + \frac{π}{2})$ $i = I_m sin(ωt + \frac{π}{2})$ - **Relationship between voltage and current:** From the formulas for voltage and current, it can be seen that in a pure capacitive circuit, the current leads the voltage by 90°. - **Diagram:** Sine wave of current leads the sine wave of voltage by 90 degrees. # Capacitive reactance The capacitive reactance Xc is given as: $X_c = \frac{1}{ωC} = \frac{1}{2πfC}$ # Instantaneous power The instantaneous power in the circuit can be derived as follows: $p = vi$ $p = (V_m sin(ωt))(I_m sin(ωt + \frac{π}{2}))$ $p = V_m I_m sin(ωt)cos(ωt)$ $p = \frac{V_m I_m}{2} sin(2ωt)$ - **Nature of power:** The instantaneous power is fluctuating in nature. # Average power The average power over one cycle can be found as follows: $P = \frac{1}{2π} \int_{0}^{2π} [\frac{V_m I_m}{2} sin(2ωt)]dωt$ $P = 0$ - **Power consumed:** The average power in a pure capacitive circuit is zero. In other words, the power consumed by a pure capacitance is zero. # Series Resistance-Inductance (R-L) Circuit - **Diagram:** A circuit diagram shows a resistor of resistance R ohms and a purely inductive coil of inductance L henry in series. - **Voltage and current:** In a series circuit, the current flowing through both R and L will be the same, but the voltage across them will be different. The vector sum of the voltage across resistor VR and the voltage across inductor VL will be equal to the supply voltage vt. - **Diagram:** A graph showing the waveforms of voltage and current in a series R-L circuit. - **Phasor Diagram:** A phasor diagram shows the current, the voltage across the resistor VR, the voltage across the inductor VL and the supply voltage V. - **Voltage Triangle:** The voltage triangle represents the voltage across the resistor VR, the voltage across the inductor VL and the supply voltage V. - **Impedance Triangle:** The impedance triangle represents the resistance R, the inductive reactance XL and the impedance Z. - **Power Triangle:** The power triangle represents the real power P, the reactive power Q and the apparent power S. - **Power:** The instantaneous value of power drawn by this circuit is given by the product of the instantaneous values of voltage and current. - **Average Power:** The average power in the circuit can be found as follows: $P_{ave} = VI cos(φ)$ # Series Resistance-Capacitance Circuit - **Diagram:** A circuit diagram shows a resistor of resistance R ohms and a purely capacitive of capacitance C farads in series. - **Voltage and current:** In a series circuit, the current flowing through both R and C will be the same, but the voltage across them will be different. The vector sum of the voltage across resistor VR and the voltage across the capacitor Vc will be equal to the supply voltage Vt. - **Diagram:** A graph showing the waveforms of voltage and current in a series R-C circuit. - **Phasor Diagram:** A phasor diagram shows the current, the voltage across the resistor VR, the voltage across the capacitor VC and the supply voltage V. - **Voltage Triangle:** The voltage triangle represents the voltage across the resistor VR, the voltage across the capacitor VC and the supply voltage V. - **Impedance Triangle:** The impedance triangle represents the resistance R, the capacitive reactance Xc and the impedance Z. - **Power Triangle:** The power triangle represents the real power P, the reactive power Q and the apparent power S. - **Power:** The instantaneous value of power drawn by this circuit is given by the product of the instantaneous values of voltage and current. - **Average Power:** The average power in the circuit can be found as follows: $P_{ave} = VI cos(ф)$ # Series Resonance RLC circuit - **Diagram:** A circuit diagram shows a resistor of resistance R ohms, a pure inductor of inductance L henry and a pure capacitor of capacitance C farads connected in series. - **Frequency:** The frequency can be varied from zero, increasing and approaching infinity. Since XL and XC are a function of frequency, at a particular frequency of applied voltage, XL and Xc will become equal in magnitude and the power factor will become unity. - **Resonance:** The circuit, when XL = XC and hence Z = R, is said to be in resonance. Since the current I remains the same throughout we can write: $I X_L$ = $I X_C$ $V_L$ = $V_C$ - **Phasor Diagram:** A phasor diagram shows that at resonance, the voltages across the inductor and capacitor will cancel out. The supply voltage will drop across the resistor R. - **Resonance Frequency:** $f_r = \frac{1}{2π√LC}$ # Q-Factor The Q- factor is nothing but the voltage magnification during resonance. # Impedance (Z) and Current (I) at Resonance - Impedance is minimum. - $Z = R$ - Current is maximum. - $I = \frac{V}{R}$ # Power factor at Resonance - Power factor is unity. - $cos(φ) = 1$ # Graphical Representation of Series Resonance RLC Circuit This circuit is shown as a graph with frequency on the x-axis. - Resistance (R) is independent of the frequency: represented by a straight line. - Inductive reactance (XL) is directly proportional to frequency: represented by a straight line. - Capacitive reactance (XC) is inversely proportional to frequency: represented by a hyperbolic curve. # Parallel Resonance RLC Circuit - **Diagram:** A parallel circuit consisting of an inductive coil with internal resistance R ohm and inductance L henry in parallel with capacitor C farads. - **Current:** The current Ic can be resolved into its active and reactive components. Its active component is I_L cosф and its reactive component is I_L sinф. # Resonance Frequency The resonance frequency is the frequency at which the power factor of the circuit becomes unity. This happens when the resultant current I is in phase with the resultant voltage V, and hence the phase angle between them is zero. - **Resonance frequency:** $f_r = \frac{1}{2π√LC} * \sqrt\frac{L}{R^2C + L}$ # Impedance At resonance, the supply current I will be in phase with the supply voltage V: $Z = \frac{L}{RC}$ - **Impedance at resonance:** The impedance during parallel resonance is very large because L and C has a very large value at that time. # Q- Factor - **Q- Factor:** - $Q = \frac{1}{RC} * \sqrt\frac{L}{R^2C + L}$ # Graphical Representation of Parallel Resonance RLC Circuit This circuit is shown as a graph with frequency on the x-axis. - Conductance (G) is represented by a straight line. - Inductive Susceptance (BL) is inversely proportional to frequency: represented by a straight line. - Capacitive Susceptance (BC) is directly proportional to frequency: represented by a straight line. # Admittance (Y) and Current (I) at Resonance - Admittance is minimum. - $Y = G$ - Current is minimum. - $I = V * G$ # Power factor at Resonance Power factor is unity. $cos(φ) = 1$ # Comparison of Series and Parallel Resonance | Sr No | Description | Series Circuit | Parallel Circuit | |---|---|---|---| | 1 | Impedance at resonance | Minimum $Z = R$ | Maximum $Z = \frac{L}{RC}$| | 2 | Current | Maximum $I = \frac{V}{R}$ | Minimum $I = \frac{V}{L/RC}$| | 3 | Resonance Frequency | $f_r = \frac{1}{2π√LC}$ | $f_r = \frac{1}{2π√LC} * \sqrt\frac{L}{R^2C + L}$| | 4 | Power Factor | Unity | Unity | | 5 | Q- Factor | $\frac{1}{RC} * \sqrt\frac{L}{R^2C + L}$ | $\frac{1}{RC} * \sqrt\frac{L}{R^2C + L}$| | 6 | It magnifies at resonance | Voltage | Current | # Power In an AC circuit, the various powers can be classified as: * Real or Active Power * Reactive Power * Apparent Power # Real Power It is denoted as P. $P = V * I * cosф = I^2R$ # Reactive Power It is denoted by Q. $Q = V * I * sinф = I^2X_L$ # Apparent Power It is denoted by S. $S = V * I = I^2Z$ $S = \sqrt{P^2 + Q^2}$ # Power Triangle The power triangle can be derived from the impedance triangle. # Three Phase Systems - **Phase Voltage:** It is defined as the voltage across either phase winding or load terminal. It is denoted by Vph. - **Line Voltage:** It is defined as the voltage across any two-line terminal. It is denoted by VL. - **Phase Current:** It is defined as the current flowing through each phase winding or load. It is denoted by Iph. - **Line Current:** It is defined as the current flowing through each line conductor. It is denoted by IL. - **Phase Sequence:** The order in which three coil emfs or currents attain their peak values is called the phase sequence. It is customary to denote the 3 phases by the three colours. i.e. red (R), yellow (Y), blue (B). - **Balanced System:** A system is said to be balanced if the voltages and currents in all phases are equal in magnitude and displaced from each other by equal angles. - **Unbalanced System:** A system is said to be unbalanced if the voltages and currents in all phases are unequal in magnitude and displaced from each other by unequal angles. - **Balanced Load:** In this type of load, all phases are equal in magnitude. This means the load will have the same power factor, and equal currents in them. # Relation between line and phase values for voltage and current in case of balanced delta connection - **Diagram:** A delta connection circuit showing the phase voltages, phase currents, and line voltages. - **Line Voltage:** $V_L = V_{ph}$ - **Line Current:** $I_L = √3 *I_{ph}$ - **Relationship between line current and phase current:** $I_L$ is the vector difference of two of the phase currents. - **Diagram:** A phasor diagram showing the line currents and phase currents. - **Power:** $P = √3 *V_L * I_L * cos(φ)$ # Relation between line and phase values for voltage and current in case of balanced star connection - **Diagram:** A star connection circuit showing the phase voltages, phase currents, and line voltages. - **Line Voltage:** $V_L = √3 *V_{ph}$ - **Line Current:** $I_L = I_{ph}$ - **Relationship between line current and phase current:** $I_L$ = $I_{ph}$ - **Diagram:** A phasor diagram showing the line voltages and phase voltages. - **Power:** $P = √3 *V_L * I_L * cos(φ)$

Use Quizgecko on...
Browser
Browser