ELECTRIC CIRCUIT ANALYSIS - I

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ELECTRIC CIRCUIT ANALYSIS - I. Chapter 15 – Series &amp; Parallel ac Circuits Lecture 19 by Moeen Ghiyas. TODAY’S lesson. Chapter 15 – Series &amp; Parallel ac Circuits. Today’s Lesson Contents. (Series ac Circuits) Impedance and Phasors Diagram Series Configuration.

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### ELECTRIC CIRCUIT ANALYSIS - I

Chapter 15 – Series & Parallel ac Circuits

Lecture 19

by MoeenGhiyas

TODAY’S lesson

Chapter 15 – Series & Parallel ac Circuits

Today’s Lesson Contents
• (Series ac Circuits)
• Impedance and Phasors Diagram
• Series Configuration
IMPEDANCE AND THE PHASOR DIAGRAM
• Resistive Elements - For the purely resistive circuit,
• Time domain equations: v = Vm sin ωt and i = Im sin ωt
• In phasor form:
• Where V = 0.707Vm and where I = 0.707Im
• Applying Ohm’s law and using phasor algebra, we have
• Since i and v are in phase, thus, θR = 0°, if phase is to be same.
• Thus, we define a new term, ZR as impedance of a resistive element (which impedes flow of current)
IMPEDANCE AND THE PHASOR DIAGRAM
• Inductive Reactance - For the inductive circuit,
• Time domain equations: v = Vm sin ωt and i = Im sin ωt
• In phasor form:
• Where V = 0.707Vm and where I = 0.707Im
• Applying Ohm’s law and using phasor algebra, we have
• Since i lags v by 90°, thus, θL = 90°, for condition to be true.
• Thus, we define term, ZL as impedance of an inductive element (which impedes flow of current)
IMPEDANCE AND THE PHASOR DIAGRAM
• Capacitive Reactance - For a capacitive circuit,
• Time domain equations: v = Vm sin ωt and i = Im sin ωt
• In phasor form:
• Where V = 0.707Vm and where I = 0.707Im
• Applying Ohm’s law and using phasor algebra, we have
• Since i leads v by 90°, thus, θC = –90°, for condition to be true.
• Thus, we define term, ZC as impedance of a capacitive element (which impedes flow of current)
IMPEDANCE AND THE PHASOR DIAGRAM
• However, it is important to realize that ZR is not a phasor, even though the format is very similar to the phasor notations for sinusoidal currents and voltages.
• The term phasor is basically reserved for quantities that vary with time, whereas R and its associated angle of 0° are fixed, i.e. non-varying quantities.
• Similarly ZL and ZC are also not phasor quantities
IMPEDANCE AND THE PHASOR DIAGRAM
• Example – Find the current i for the circuit of fig. Sketch the waveforms of v and i.
• Solution:
• In phasor form
• From ohm’s law
• Converting to time domain
IMPEDANCE AND THE PHASOR DIAGRAM
• Sketch of waveform and Phasor Diagram
IMPEDANCE AND THE PHASOR DIAGRAM
• Example – Find the voltage v for the circuit of fig. Sketch the waveforms of v and i.
• Solution:
• In phasor form
• From ohm’s law
• Converting to time domain
IMPEDANCE AND THE PHASOR DIAGRAM
• Sketch of waveform and Phasor Diagram
IMPEDANCE AND THE PHASOR DIAGRAM
• Example – Find the voltage v for the circuit of fig. Sketch the waveforms of v and i.
• Solution:
• In phasor form
• From ohm’s law
• Converting to time domain
IMPEDANCE AND THE PHASOR DIAGRAM
• Sketch of waveform and Phasor Diagram
IMPEDANCE AND THE PHASOR DIAGRAM
• Impedance Diagram - For any network,
• Resistance is plotted on the positive real axis,
• Inductive reactance on the positive imaginary axis, and
• Capacitive reactance on the negative imaginary axis.
• Impedance diagram reflects the individual and total impedance levels of ac network.
IMPEDANCE AND THE PHASOR DIAGRAM
• Impedance Diagram
• The magnitude of total impedance of a network defines the resulting current level (through Ohm’s law)
• For any configuration (series, parallel, series-parallel, etc.), the angle associated with the total impedance is the angle by which the applied voltage leads the source current.
• Thus angle of impedance reveals whether the network is primarily inductive or capacitive or simply resistive.
• For inductive networks θT will be positive, whereas for capacitive networks θT will be negative, and θT will be zero for resistive cct.
SERIES CONFIGURATION
• Overall properties of series ac circuits are the same as those for dc circuits
• For instance, the total impedance of a system is the sum of the individual impedances:
SERIES CONFIGURATION
• EXAMPLE - Determine the input impedance to the series network of fig. Draw the impedance diagram.
• Solution:
SERIES CONFIGURATION
• EXAMPLE - Determine the input impedance to the series network of fig. Draw the impedance diagram.
• Solution:
SERIES CONFIGURATION
• Current is same in ac series circuits just like it is in dc circuits.
• Ohm’s law applicability is same.
• KVL applies in similar manner.
• The power to the circuit can be determined by
• where θT is the phase angle between E and I.
SERIES CONFIGURATION
• Impedance Relation with Power Factor
• We know that
• Reference to figs and equations
• θT is not only the impedance angle of ZT but also θT is the phase angle between the input voltage and current for a series ac circuit.

Impedance Diagram

Phasor Diagram

Note: θT of ZT is with reference to voltage unlike FP . Also current I is in phase with VR, lags the VL by 90°, and leads the VC by 90°.

SERIES CONFIGURATION
• R-L-C Example
• Step 1 – Convert Available information to Phasor Notation
SERIES CONFIGURATION
• R-L-C Example
• . Step 2 – Find ZT and make impedance diagram
SERIES CONFIGURATION
• R-L-C Example
• Step 3 – Find I or E
SERIES CONFIGURATION
• R-L-C Example
• Step 4 – Find phasor voltages across each element
SERIES CONFIGURATION
• R-L-C Example
• I =
• VR =
• VL =
• VC =
• . Step 5 – Make phasor diagram and
• . apply KVL (for verification or if req)

Note: Current I in phase with VR, lags the VL by 90°, and leads the VC by 90°

SERIES CONFIGURATION
• R-L-C Example
• Step 6 – Convert phasor values to time domain
SERIES CONFIGURATION
• R-L-C Example
• Step 7 – Plot all the voltages and the current of the circuit
SERIES CONFIGURATION
• R-L-C Example
• Step 8 – Calculation of total power in watts delivered to the circuit
• or
• or
SERIES CONFIGURATION
• R-L-C Example
• Step 9 – The power factor of the circuit is
• or
Summary / Conclusion
• (Series ac Circuits)
• Impedance and Phasors Diagram
• Series Configuration