12P07��Alternating Current
12P07.1
� AC Voltage Applied to a Resistor
12P07.1 AC Voltage Applied to a Resistor
Learning Objectives
What is AC
AC voltage applied to a purely Resistive Circuit
Power developed in a purely Resistive Circuit
Representation of AC Current and Voltage by Phasors
12P07.1
CV1
What is AC
Types of Current
What is AC
Types of Current
Direct Current
What is AC
Types of Current
Direct Current
Alternating Current
What is AC
Types of Current
Direct Current
Alternating Current
What is AC
Current does not change direction with time
Types of Current
Direct Current
Alternating Current
What is AC
Current which changes direction at regular intervals with time
What is AC
This symbol represents the AC
What is AC
Generally AC is represented by sine curves
What is AC
ω is the angular frequency of voltage
What is AC
vm is Amplitude or peak value of voltage
What is AC
Do you know why we prefer AC over DC ?
What is AC
Advantages of AC over DC
1. Easily and efficiently convertible from one voltage level to other voltage level.
Blue ( #307bf3ff) boxes with white text
What is AC
Advantages of AC over DC
2. Economic transmission of electrical energy over long distance.
Blue ( #307bf3ff) boxes with white text
What is AC
ConcepTest
Ready for challenge
What is AC
Q. In the following images, classify the currents as AC or non AC?
(i)
(ii)
(iii)
Pause the Video
(Time Duration : 1 Minute)
What is AC
Sol.
(i)
Non AC, as the current is not changing it’s direction.
Only magnitude of the current is changing.
What is AC
Sol.
(ii)
AC, as the current is changing it’s direction with time and also the amplitude (a) and time duration (2 t1) for each cycle is same.
What is AC
Sol.
(iii)
Non AC, as the current is not changing it’s direction.
Only magnitude of the current is changing.
12P07.1
CV2
AC Voltage Applied to a purely Resistive Circuit
AC Voltage Applied to a purely Resistive Circuit
About circuit
Resistor R
AC Voltage Applied to a purely Resistive Circuit
About circuit
Resistor R
Instantaneous voltage across R
AC Voltage Applied to a purely Resistive Circuit
About circuit
Resistor R
Instantaneous voltage across R
Instantaneous voltage of voltage source
AC Voltage Applied to a purely Resistive Circuit
About circuit
Resistor R
Instantaneous voltage across R
Instantaneous voltage of voltage source
Maximum voltage of the voltage source
AC Voltage Applied to a purely Resistive Circuit
Mathematical Analysis of the Circuit
By kirchhoff’s law
∑ v(t) = 0
AC Voltage Applied to a purely Resistive Circuit
Mathematical Analysis of the Circuit
By kirchhoff’s law
∑ v(t) = 0
ie. v - vR = 0
AC Voltage Applied to a purely Resistive Circuit
Mathematical Analysis of the Circuit
By kirchhoff’s law
∑ v(t) = 0
ie. v - vR = 0
Therefore, vm sin ω t = i R
AC Voltage Applied to a purely Resistive Circuit
Mathematical Analysis of the Circuit
By kirchhoff’s law
∑ v(t) = 0
ie. v - vR = 0
Therefore, vm sin ω t = i R
or i = (vm sin ω t / R)
AC Voltage Applied to a purely Resistive Circuit
Mathematical Analysis of the Circuit
By kirchhoff’s law
∑ v(t) = 0
ie. v - vR = 0
Therefore, vm sin ω t = i R
or i = (vm sin ω t / R)
or i = im sin ωt
AC Voltage Applied to a purely Resistive Circuit
Mathematical Analysis of the Circuit
By kirchhoff’s law
∑ v(t) = 0
ie. v - vR = 0
Therefore, vm sin ω t = iR
or i = (vm sin ω t / R)
or i = im sin ω t
where im= (vm / R)
AC Voltage Applied to a purely Resistive Circuit
Mathematical Analysis of the Circuit
On comparing v = vm sin ωt and i = im sin ωt
Like voltage, the current also varies sinusoidally
AC Voltage Applied to a purely Resistive Circuit
Mathematical Analysis of the Circuit
Voltage and current both reach their maximum and minimum values at the same time.
That means they are in same phase
AC Voltage Applied to a purely Resistive Circuit
Mathematical Analysis of the Circuit
The sum of the instantaneous current values over one complete cycle is zero
AC Voltage Applied to a purely Resistive Circuit
Mathematical Analysis of the Circuit
The average current is also zero
12P07.1
CV3
Power developed in a purely Resistive Circuit
Power developed in a purely Resistive Circuit
As the average current is zero.
Does that mean average power is also zero ?
Power developed in a purely Resistive Circuit
The instantaneous power dissipated in the resistor
Power developed in a purely Resistive Circuit
The instantaneous power dissipated in the resistor
The average value of power over a cycle
Power developed in a purely Resistive Circuit
The instantaneous power dissipated in the resistor
The average value of power over a cycle
or
Power developed in a purely Resistive Circuit
The instantaneous power dissipated in the resistor
The average value of power over a cycle
or or
Power developed in a purely Resistive Circuit
The instantaneous power dissipated in the resistor
The average value of power over a cycle
or or
or
Power developed in a purely Resistive Circuit
The instantaneous power dissipated in the resistor
The average value of power over a cycle
or or
or or
Power developed in a purely Resistive Circuit
For a complete cycle
Power developed in a purely Resistive Circuit
For a complete cycle
( AC Power )
Power developed in a purely Resistive Circuit
For a complete cycle
( AC Power )
P = I2 R ( DC Power )
Power developed in a purely Resistive Circuit
For a complete cycle
( AC Power )
P = I2 R ( DC Power )
To express AC power in the same form as DC power, Root Mean Square (RMS) current (Irms or I) is defined.
Power developed in a purely Resistive Circuit
For a complete cycle
( AC Power )
P = I2 R ( DC Power )
To express AC power in the same form as DC power, Root Mean Square (RMS) current (Irms or I) is defined.
So,
Power developed in a purely Resistive Circuit
For a complete cycle
( AC Power )
P = I2 R ( DC Power )
To express AC power in the same form as DC power, Root Mean Square (RMS) current (Irms or I) is defined.
So,
Power developed in a purely Resistive Circuit
For a complete cycle
( AC Power )
P = I2 R ( DC Power )
To express AC power in the same form as DC power, Root Mean Square (RMS) current (Irms or I) is defined.
So,
Similarly,
PSV 1
Q. If RMS value of voltage is 250 V, then find its peak value.
Q. If RMS value of voltage is 250 V, then find its peak value.
Sol. We know Vrms = 0.707 vm
Q. If RMS value of voltage is 250 V, then find its peak value.
Sol. We know Vrms = 0.707 vm
⇒ vm = (Vrms / 0.707)
Q. If RMS value of voltage is 250 V, then find its peak value.
Sol. We know Vrms = 0.707 vm
⇒ vm = (Vrms / 0.707)
= (250 / 0.707)
Q. If RMS value of voltage is 250 V, then find its peak value.
Sol. We know Vrms = 0.707 vm
⇒ vm = (Vrms / 0.707)
= (250 / 0.707)
= 353.61 Volt
PSV 2
Q. An electric heater is rated as 1000 W for a 220 V supply. Find
Q. An electric heater is rated as 1000 W for a 220 V supply. Find
(i) Resistance of the heater.
Q. An electric heater is rated as 1000 W for a 220 V supply. Find
(i) Resistance of the heater.
(ii) Peak value of the voltage.
Q. An electric heater is rated as 1000 W for a 220 V supply. Find
(i) Resistance of the heater.
(ii) Peak value of the voltage.
(iii) RMS value of the current.
Pause the Video
(Time Duration : 3 Minutes)
Q. An electric heater is rated as 1000 W for a 220 V supply. Find
(i) Resistance of the heater.
(ii) Peak value of the voltage.
(iii) RMS value of the current.
Sol. (i) For resistance of the heater
∵ P = (I2 R) = (V2 / R)
Q. An electric heater is rated as 1000 W for a 220 V supply. Find
(i) Resistance of the heater.
(ii) Peak value of the voltage.
(iii) RMS value of the current.
Sol. (i) For resistance of the heater
∵ P = (I2 R) = (V2/ R)
∴ R = (V2 / P)
Q. An electric heater is rated as 1000 W for a 220 V supply. Find
(i) Resistance of the heater.
(ii) Peak value of the voltage.
(iii) RMS value of the current.
Sol. (i) For resistance of the heater
∵ P = (I2 R) = (V2 / R)
∴ R = (V2 / P)
= (2202 / 1000) = 48.4 Ω
Sol.
(ii) For peak value of the voltage
∵ Vrms = 0.707 vm
Sol.
(ii) For peak value of the voltage
∵ Vrms = 0.707 vm
∴ vm = (Vrms / 0.707)
Sol.
(ii) For peak value of the voltage
∵ Vrms = 0.707 vm
∴ vm = (Vrms / 0.707)
= (220 / 0.707) = 311.17 Volt
Sol.
(ii) For peak value of the voltage
∵ Vrms = 0.707 vm
∴ vm = (Vrms / 0.707)
= (220 / 0.707) = 311.17 Volt
(iii) For RMS value of the current
∵ P = I V
Sol.
(ii) For peak value of the voltage
∵ Vrms = 0.707 vm
∴ vm = (Vrms / 0.707)
= (220 / 0.707) = 311.17 Volt
(iii) For RMS value of the current
∵ P = I V
∴ I = (P / V)
Sol.
(ii) For peak value of the voltage
∵ Vrms = 0.707 vm
∴ vm = (Vrms / 0.707)
= (220 / 0.707) = 311.17 Volt
(iii) For RMS value of the current
∵ P = I V
∴ I = (P / V)
= (1000/ 220) = 4.54 Ampere
12P07.1
CV4
Representation of AC Current and Voltage by Phasors
Representation of AC Current and Voltage by Phasors
Phasor and Phasor Diagram
Phasor is a vector whose length is the amplitude or peak value of voltage or current
Representation of AC Current and Voltage by Phasors
Phasor and Phasor Diagram
Phasor rotates counterclockwise about the origin with an angular speed ω
Representation of AC Current and Voltage by Phasors
Phasor and Phasor Diagram
The projection of the phasor onto the vertical axis represents the instantaneous value of the voltage or current
Representation of AC Current and Voltage by Phasors
Phasor and Phasor Diagram
The projection of the phasor onto the vertical axis represents the instantaneous value of the voltage or current
Representation of AC Current and Voltage by Phasors
Phasor and Phasor Diagram
The projection of the phasor onto the vertical axis represents the instantaneous value of the voltage or current
Phasor Diagram for a purely Resistive Circuit
Phasor and Phasor Diagram
This phasor diagram is for a purely resistive circuit
Phasor Diagram for a purely Resistive Circuit
Phasor and Phasor Diagram
Phasors V and I are in same direction and this is for all the times
Phasor Diagram for a purely Resistive Circuit
Phasor and Phasor Diagram
For a purely resistive circuit the phase angle between the voltage and the current is zero
PSV 3
Q. An electric bulb reads 200 W and 220 V. Find the
Q. An electric bulb reads 200 W and 220 V. Find the
Q. An electric bulb reads 200 W and 220 V. Find the
Pause the Video
(Time Duration : 2 Minutes)
Q. An electric bulb reads 200 W and 220 V. Find the
Sol.
R = (V2 / P)
Q. An electric bulb reads 200 W and 220 V. Find the
Sol.
R = (V2 / P)
= (220 × 220 / 200) = 242 Ω
Q. An electric bulb reads 200 W and 220 V. Find the
Sol.
R = (V2 / P)
= (220 × 220 / 200) = 242 Ω
(ii) RMS value of the current
I = (P / V)
Q. An electric bulb reads 200 W and 220 V. Find the
Sol.
R = (V2 / P)
= (220 × 220 / 200) = 242 Ω
(ii) RMS value of the current
I = (P / V)
= (200 / 220) ≈ 0.91 A
Summary
12P07.1 AC Voltage Applied to a Resistor
Reference Questions
NCERT : 7.1, 7.2
Work Book : 10
12P07.2
AC Voltage Applied to an Inductor
Learning Objective
AC Voltage Applied to a purely Inductive Circuit
Phasor Diagram For a Purely Inductive Circuit
Power in a Purely Inductive Circuit
Magnetisation and Demagnetisation of an Inductor
12P07.2 AC Voltage Applied to an Inductor
12P07.2
CV1
AC Voltage Applied to a purely Inductive Circuit
AC Voltage Applied to a purely Inductive Circuit
About circuit
Inductor L
AC Voltage Applied to a purely Inductive Circuit
About circuit
Inductor L
Instantaneous voltage across L
AC Voltage Applied to a purely Inductive Circuit
About circuit
Inductor L
Instantaneous voltage across L
Instantaneous voltage of voltage source
AC Voltage Applied to a purely Inductive Circuit
About circuit
Inductor L
Instantaneous voltage across L
Instantaneous voltage of voltage source
Maximum voltage of the voltage source
AC Voltage Applied to a purely Inductive Circuit
Mathematical Analysis of Circuit
On applying Kirchhoff’s loop rule,
∑ v(t) = 0
AC Voltage Applied to a purely Inductive Circuit
Mathematical Analysis of Circuit
On applying Kirchhoff’s loop rule,
∑ v(t) = 0
Therefore, v - vL = 0
AC Voltage Applied to a purely Inductive Circuit
Mathematical Analysis of Circuit
On applying Kirchhoff’s loop rule,
∑ v(t) = 0
Therefore, v - vL = 0
or vm sin ωt = L(di / dt)
AC Voltage Applied to a purely Inductive Circuit
Mathematical Analysis of Circuit
On applying Kirchhoff’s loop rule,
∑ v(t) = 0
Therefore, v - vL = 0
or vm sinωt = L(di /dt)
or (di / dt) = (vm / L) sin ωt
AC Voltage Applied to a purely Inductive Circuit
Mathematical Analysis of Circuit
On applying Kirchhoff’s loop rule,
∑ v(t) = 0
Therefore, v - vL = 0
or vm sin ωt = L(di / dt)
or (di / dt) = (vm / L) sinωt
To obtain the current,
AC Voltage Applied to a purely Inductive Circuit
Mathematical Analysis of Circuit
On applying Kirchhoff’s loop rule,
∑ v(t) = 0
Therefore, v - vL = 0
or vm sin ωt = L(di / dt)
or (di / dt) = (vm / L) sinωt
To obtain the current,
⇒ i = (-vm / ωL) cos(ωt) + constant
AC Voltage Applied to a purely Inductive Circuit
The integration constant is zero.
AC Voltage Applied to a purely Inductive Circuit
The integration constant is zero.
And also
AC Voltage Applied to a purely Inductive Circuit
The integration constant is zero.
And also
So,
AC Voltage Applied to a purely Inductive Circuit
The integration constant is zero.
And also
So,
where im = (vm / ωL)
AC Voltage Applied to a purely Inductive Circuit
The integration constant is zero.
And also
So,
where im = (vm / ωL)
XL = ωL (inductive reactance)
AC Voltage Applied to a purely Inductive Circuit
The integration constant is zero.
And also
So,
where im = (vm / ωL)
XL = ωL (inductive reactance)
So, im = (vm / XL)
AC Voltage Applied to a purely Inductive Circuit
The integration constant is zero.
And also
So,
where im = (vm / ωL)
XL = ωL (inductive reactance)
So, im = (vm / XL)
The SI unit of XL is ohm (Ω).
AC Voltage Applied to a purely Inductive Circuit
The integration constant is zero.
And also
So,
where im = (vm / ωL)
XL = ωL (inductive reactance)
So, im = (vm / XL)
The SI unit of XL is ohm (Ω).
XL ∝ ω and XL ∝ L
PSV 4
Q. A pure inductor of 25.0 mH is connected to a source of 220 V. Find the
if the frequency of the source is 50 Hz.
Q. A pure inductor of 25.0 mH is connected to a source of 220 V. Find the
if the frequency of the source is 50 Hz.
Sol.
XL = ωL
Q. A pure inductor of 25.0 mH is connected to a source of 220 V. Find the
if the frequency of the source is 50 Hz.
Sol.
XL = ωL
= 2 π f L
Q. A pure inductor of 25.0 mH is connected to a source of 220 V. Find the
if the frequency of the source is 50 Hz.
Sol.
XL = ωL
= 2 π f L
= 2 × 3.14 × 50 × 25 × 10-3
= 7.85 Ω
Q. A pure inductor of 25.0 mH is connected to a source of 220 V. Find the
if the frequency of the source is 50 Hz.
Sol.
XL = ωL
= 2 π f L
= 2 × 3.14 × 50 × 25 × 10-3
= 7.85 Ω
(ii) The RMS current in the circuit is
I = (V / XL)
Q. A pure inductor of 25.0 mH is connected to a source of 220 V. Find the
if the frequency of the source is 50 Hz.
Sol.
XL = ωL
= 2 π f L
= 2 × 3.14 × 50 × 25 × 10-3
= 7.85 Ω
(ii) The RMS current in the circuit is
I = (V / XL)
= (220 V / 7.85 Ω)
Q. A pure inductor of 25.0 mH is connected to a source of 220 V. Find the
if the frequency of the source is 50 Hz.
Sol.
XL = ωL
= 2 π f L
= 2 × 3.14 × 50 × 25 × 10-3
= 7.85 Ω
(ii) The RMS current in the circuit is
I = (V / XL)
= (220 V / 7.85 Ω)
= 28 A
12P07.2
CV2
Phasor Diagram For a Purely Inductive Circuit
Phasor Diagram For a Purely Inductive Circuit
On comparing v = vm sin ωt and
The current lags the voltage by π / 2.
Phasor Diagram For a Purely Inductive Circuit
Current reaches its maximum value later than the voltage by one-fourth of a period
Phasor Diagram For a Purely Inductive Circuit
The current phasor I is π / 2 behind the voltage phasor V
Phasor Diagram For a Purely Inductive Circuit
The projection of the phasor onto the vertical axis represents the instantaneous value of the voltage or current
Phasor Diagram For a Purely Inductive Circuit
Simultaneously Phasor and Graph representation of current and voltage
12P07.2
CV3
Power in a Purely Inductive Circuit
Power in a Purely Inductive Circuit
The instantaneous power supplied to the inductor
Power in a Purely Inductive Circuit
The instantaneous power supplied to the inductor
Power in a Purely Inductive Circuit
The instantaneous power supplied to the inductor
Power in a Purely Inductive Circuit
The instantaneous power supplied to the inductor
Power in a Purely Inductive Circuit
The instantaneous power supplied to the inductor
The average power over a complete cycle
Power in a Purely Inductive Circuit
The instantaneous power supplied to the inductor
The average power over a complete cycle
Power in a Purely Inductive Circuit
The instantaneous power supplied to the inductor
The average power over a complete cycle
PSV 5
Q. If an inductor of 10 H is connected across the voltage source of v = vm sin ωt.
Find the average power in T / 4 second, where T is time period.
Sol. For an inductor average power
Sol. For an inductor average power
For T / 4 cycle
Sol. For an inductor average power
For T / 4 cycle
Sol. For an inductor average power
For T / 4 cycle
Sol. For an inductor average power
For T / 4 cycle
Sol. For an inductor average power
For T / 4 cycle
Sol. For an inductor average power
For T / 4 cycle
Sol. For an inductor average power
For T / 4 cycle
12P07.2
CV4
Magnetisation and Demagnetisation of an Inductor
Magnetisation and Demagnetisation of an Inductor
For an inductor
Current lags voltage by π / 2
Magnetisation and Demagnetisation of an Inductor
For an inductor
flux (Φ) = B A
and B ∝ i
⇒ Φ ∝ i
Magnetisation and Demagnetisation of an Inductor
For (0 - 1)
v ⇒ +ve
i ⇒ +ve ⇒ Energy is absorbed from the source
p ⇒ +ve
Magnetisation and Demagnetisation of an Inductor
For (1 - 2)
v ⇒ - ve
i ⇒ +ve ⇒ Energy is being returned to the source
p ⇒ - ve
Magnetisation and Demagnetisation of an Inductor
For (2 - 3)
v ⇒ - ve
i ⇒ - ve ⇒ Energy is absorbed from the source
p ⇒ +ve
Magnetisation and Demagnetisation of an Inductor
For (3 - 4)
v ⇒ +ve
i ⇒ - ve ⇒ Energy is being returned to the source
p ⇒ - ve
PSV 6
Q. An inductor of inductance L = 5 H is connected to an AC source of voltage
V(t) = 10 sin [10t + (π / 6)]. Find the
Q. An inductor of inductance L = 5 H is connected to an AC source of voltage
V(t) = 10 sin [10t +(π / 6)]. Find the
Q. An inductor of inductance L = 5 H is connected to an AC source of voltage
V(t) = 10 sin [10t +(π / 6)]. Find the
Q. An inductor of inductance L = 5 H is connected to an AC source of voltage
V(t) = 10 sin [10t +(π / 6)]. Find the
Q. An inductor of inductance L = 5 H is connected to an AC source of voltage
V(t) = 10 sin [10t +(π / 6)]. Find the
Sol. (i) Inductive reactance
XL = ωL
Q. An inductor of inductance L = 5 H is connected to an AC source of voltage
V(t) = 10 sin [10t +(π / 6)]. Find the
Sol. (i) Inductive reactance
XL = ωL
= 10 × 5
= 50 Ω
Q. An inductor of inductance L = 5 H is connected to an AC source of voltage
V(t) = 10 sin [10t +(π / 6)]. Find the
Sol. (i) Inductive reactance
XL = ωL
= 10 × 5
= 50 Ω
(ii) Peak voltage (Vm) = 10 Volt
Q. An inductor of inductance L = 5 H is connected to an AC source of voltage
V(t) = 10 sin [10t +(π / 6)]. Find the
Sol. (i) Inductive reactance
XL = ωL
= 10 × 5
= 50 Ω
(ii) Peak voltage (Vm) = 10 Volt
RMS voltage (Vrms) = (10 / √2) Volt
Sol. (iii) Peak current (im) = (Vm / XL)
Sol. (iii) Peak current (im) = (Vm / XL)
= (10 / 50)
= (1 / 5) A
Sol. (iii) Peak current (im) = (Vm / XL)
= (10 / 50)
= (1 / 5) A
RMS current (irms) = (im / √2)
Sol. (iii) Peak current (im) = (Vm / XL)
= (10 / 50)
= (1 / 5) A
RMS current (irms) = (im / √2)
= (1 / 5 √2) A
Summary
12P07.2 AC Voltage Applied to an Inductor
Reference Questions
NCERT : 7.3
Work Book : 10
12P07.3
AC Voltage Applied to a Capacitor
AC Voltage Applied to a Capacitor
Learning Objective
AC Voltage applied to a Purely Capacitive Circuit
Phasor Diagram For a Purely Capacitive Circuit
Power in a Purely Capacitive Circuit
Charging and Discharging of a Capacitor
12P07.3
CV1
AC Voltage applied to a Purely Capacitive Circuit
AC Voltage Applied to a purely capacitive Circuit
About circuit
Capacitor C
AC Voltage Applied to a purely capacitive Circuit
About circuit
Capacitor C
Instantaneous voltage (vC = (q / C)) across C
AC Voltage Applied to a purely capacitive Circuit
About circuit
Capacitor C
Instantaneous voltage (vC = (q / C)) across C
Instantaneous voltage of voltage source
AC Voltage Applied to a purely capacitive Circuit
About circuit
Capacitor C
Instantaneous voltage (vC = (q / C)) across C
Instantaneous voltage of voltage source
Maximum voltage of the voltage source
AC Voltage Applied to a purely capacitive Circuit
Mathematical Analysis of Circuit
On applying Kirchhoff’s loop rule,
∑ v(t) = 0
AC Voltage Applied to a purely capacitive Circuit
Mathematical Analysis of Circuit
On applying Kirchhoff’s loop rule,
∑ v(t) = 0
Therefore, v - vC= 0
AC Voltage Applied to a purely capacitive Circuit
Mathematical Analysis of Circuit
On applying Kirchhoff’s loop rule,
∑ v(t) = 0
Therefore, v - vC= 0
or vmsin ωt = (q / C)
AC Voltage Applied to a purely capacitive Circuit
Mathematical Analysis of Circuit
On applying Kirchhoff’s loop rule,
∑ v(t) = 0
Therefore, v - vC= 0
or vmsin ωt = (q / C)
or q = vmC sin ωt
AC Voltage Applied to a purely capacitive Circuit
Mathematical Analysis of Circuit
On applying Kirchhoff’s loop rule,
∑ v(t) = 0
Therefore, v - vC= 0
or vmsin ωt = (q / C)
or q = vmC sin ωt
To find the current,
i = (dq / dt)
AC Voltage Applied to a purely capacitive Circuit
Mathematical Analysis of Circuit
On applying Kirchhoff’s loop rule,
∑ v(t) = 0
Therefore, v - vC= 0
or vmsin ωt = (q / C)
or q = vmC sin ωt
To find the current,
i = (dq / dt)
or i = d/dt (vmC sin ωt)
AC Voltage Applied to a purely capacitive Circuit
Mathematical Analysis of Circuit
On applying Kirchhoff’s loop rule,
∑ v(t) = 0
Therefore, v - vC= 0
or vmsin ωt = (q / C)
or q = vmC sin ωt
To find the current,
i = (dq / dt)
or i = d/dt (vmC sin ωt)
i = ω C vm cos(ωt)
AC Voltage Applied to a purely capacitive Circuit
And also
AC Voltage Applied to a purely capacitive Circuit
And also
So,
AC Voltage Applied to a purely capacitive Circuit
And also
So,
where im = (ω C vm)
AC Voltage Applied to a purely capacitive Circuit
And also
So,
where im = (ω C vm)
or
AC Voltage Applied to a purely capacitive Circuit
And also
So,
where im = (ω C vm)
or
Xc = (1/ ω C)
AC Voltage Applied to a purely capacitive Circuit
And also
So,
where im = (ω C vm)
or
Xc = (1/ ω C)
So, im = (vm / Xc )
AC Voltage Applied to a purely capacitive Circuit
And also
So,
where im = (ω C vm)
or
Xc = (1/ ω C)
So, im = (vm / Xc )
The SI unit of XC is ohm (Ω).
AC Voltage Applied to a purely capacitive Circuit
And also
So,
where im = (ω C vm)
or
Xc = (1/ ω C)
So, im = (vm / Xc )
The SI unit of XC is ohm (Ω).
Xc ∝ (1/ω) and Xc ∝ (1/C)
PSV 7
Q. A 2.5 µF capacitor is connected to a fan, which is supplied by 220 V, 50 Hz source.
Find the
Q. A 2.5 µF capacitor is connected to a fan, which is supplied by 220 V, 50 Hz source.
Find the
Q. A 2.5 µF capacitor is connected to a fan, which is supplied by 220 V, 50 Hz source.
Find the
Sol. (i) The capacitive reactance
XC = (1 / 2π f C)
Q. A 2.5 µF capacitor is connected to a fan, which is supplied by 220 V, 50 Hz source.
Find the
Sol. (i) The capacitive reactance
XC = (1 / 2π f C)
= 1 / 2π (50 Hz)(2.5 X 10-6 F)
= 1273.24 Ω
Q. A 2.5 µF capacitor is connected to a fan, which is supplied by 220 V, 50 Hz source.
Find the
Sol. (i) The capacitive reactance
XC = (1 / 2π f C)
= 1 / 2π (50 Hz)(2.5 X 10-6 F)
= 1273.24 Ω
(ii) The RMS current
I = (V / XC)
Q. A 2.5 µF capacitor is connected to a fan, which is supplied by 220 V, 50 Hz source.
Find the
Sol. (i) The capacitive reactance
XC = (1 / 2π f C)
= 1 / 2π (50 Hz)(2.5 X 10-6 F)
= 1273.24 Ω
(ii) The RMS current
I = (V / XC)
= (220 V / 1273.24 Ω )
= 0.1728 A
12P07.3
CV2
Phasor Diagram For a Purely Capacitive Circuit
Phasor Diagram For a Purely Capacitive Circuit
On comparing v = vm sin ωt and
The current leads the voltage by π/2.
Phasor Diagram For a Purely Capacitive Circuit
Current reaches its maximum value before the voltage by one-fourth of a period
Phasor Diagram For a Purely Capacitive Circuit
The current phasor I is π/2 ahead of the voltage phasor V
Phasor Diagram For a Purely Capacitive Circuit
The projection of the phasor onto the vertical axis represents the instantaneous value of the voltage or current
Phasor Diagram For a Purely Capacitive Circuit
Simultaneously Phasor and Graph representation of current and voltage
12P07.3
CV3
Power in a Purely Capacitive Circuit
Power in a Purely Capacitive Circuit
The instantaneous power supplied to the capacitor is
Power in a Purely Capacitive Circuit
The instantaneous power supplied to the capacitor is
Power in a Purely Capacitive Circuit
The instantaneous power supplied to the capacitor is
Power in a Purely Capacitive Circuit
The instantaneous power supplied to the capacitor is
The average power over a complete cycle is
Power in a Purely Capacitive Circuit
The instantaneous power supplied to the capacitor is
The average power over a complete cycle is
Power in a Purely Capacitive Circuit
The instantaneous power supplied to the capacitor is
The average power over a complete cycle is
12P07.3
CV4
Charging and Discharging of a Capacitor
Charging and Discharging of a Capacitor
For a capacitor
Current leads voltage by π / 2
Charging and Discharging of a Capacitor
For a capacitor
voltage vC = q / C
⇒ vC ∝ q
Charging and Discharging of a Capacitor
For (0-1)
vC ⇒ + ve
i ⇒ + ve ⇒ Energy is absorbed from the source
p ⇒ +ve
Charging and Discharging of a Capacitor
For (1-2)
v ⇒ +ve
i ⇒ - ve ⇒ Energy is being returned to the source
p ⇒ - ve
Charging and Discharging of a Capacitor
For (2-3)
v ⇒ -ve
i ⇒ - ve ⇒ Energy is absorbed from the source
p ⇒ + ve
Charging and Discharging of a Capacitor
For (3-4)
v ⇒ - ve
i ⇒ + ve ⇒ Energy is being returned to the source
p ⇒ - ve
PSV 8
Q. A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the
Q. A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the
Q. A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the
Q. A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the
Q. A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the
Sol. (i) The capacitive reactance is
XC = (1 / 2π f C)
Q. A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the
Sol. (i) The capacitive reactance is
XC = (1 / 2π f C)
= 1 / 2π (50 Hz)(15.0 X 10-6 F)
= 212 Ω
Q. A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the
Sol. (i) The capacitive reactance is
XC = (1 / 2π f C)
= 1 / 2π (50 Hz)(15.0 X 10-6 F)
= 212 Ω
(ii) The RMS current is
I = (V / XC)
Q. A 15.0 µF capacitor is connected to a 220 V, 50 Hz source. Find the
Sol. (i) The capacitive reactance is
XC = (1 / 2π f C)
= 1 / 2π (50 Hz)(15.0 X 10-6 F)
= 212 Ω
(ii) The RMS current is
I = (V / XC)
= (220 V / 212 Ω )
= 1.04 A
Sol. (iii) The peak current is
im = √2 I
Sol. (iii) The peak current is
im = √2 I
= (1.41)(1.04 A)
= 1.47 A
Sol. (iii) The peak current is
im = √2 I
= (1.41)(1.04 A)
= 1.47 A
(iv) As,
Sol. (iii) The peak current is
im = √2 I
= (1.41)(1.04 A)
= 1.47 A
(iv) As,
So, if the frequency is doubled, the capacitive reactance is halved.
Sol. (iii) The peak current is
im = √2 I
= (1.41)(1.04 A)
= 1.47 A
(iv) As,
So, if the frequency is doubled, the capacitive reactance is halved.
Now,
Sol. (iii) The peak current is
im = √2 I
= (1.41)(1.04 A)
= 1.47 A
(iv) As,
So, if the frequency is doubled, the capacitive reactance is halved.
Now,
So, as capacitive reactance is halved, the current will be doubled.
a current i = imsin(ωt + π / 2) in the capacitor, where im= vm / XC .
Summary
12P07.3 AC Voltage Applied to a Capacitor
Reference Questions
NCERT : 7.4,7.5
12P07.4
Series LCR Circuit
Series LCR Circuit
Learning Objective
Phasor-diagram Solution of Series LCR Circuit
Analytical Solution of Series LCR Circuit
Resonance
12P07.4
CV1
Phasor-diagram Solution of Series LCR Circuit
Phasor-diagram Solution of Series LCR Circuit
AC voltage source.
Phasor-diagram Solution of Series LCR Circuit
AC voltage source.
Phasor-diagram Solution of Series LCR Circuit
AC voltage source.
v L+ v R+ v C= v
Phasor-diagram Solution of Series LCR Circuit
AC voltage source.
v L+ v R+ v C= v
i = imsin(ωt + ϕ)
Phasor-diagram Solution of Series LCR Circuit
AC voltage source.
v L+ v R+ v C= v
i = imsin(ωt + ϕ)
where ϕ is phase difference.
Phasor-diagram Solution of Series LCR Circuit
The current phasor is I.
Phasor-diagram Solution of Series LCR Circuit
The current phasor is I.
VR is voltage across R and is parallel to I.
Phasor-diagram Solution of Series LCR Circuit
The current phasor is I.
VR is voltage across R and is parallel to I.
The amplitude of VR is vRm= imR.
Phasor-diagram Solution of Series LCR Circuit
The current phasor is I.
VR is voltage across R and is parallel to I.
The amplitude of VR is vRm= imR.
VL is voltage across L and is π/2 ahead of I.
Phasor-diagram Solution of Series LCR Circuit
The current phasor is I.
VR is voltage across R and is parallel to I.
The amplitude of VR is vRm= imR.
VL is voltage across L and is π/2 ahead of I.
The amplitude of VL is vLm= imXL.
Phasor-diagram Solution of Series LCR Circuit
The current phasor is I.
VR is voltage across R and is parallel to I.
The amplitude of VR is vRm= imR.
VL is voltage across L and is π/2 ahead of I.
The amplitude of VL is vLm= imXL.
VC is voltage across C and is π/2 behind I.
Phasor-diagram Solution of Series LCR Circuit
The current phasor is I.
VR is voltage across R and is parallel to I.
The amplitude of VR is vRm= imR.
VL is voltage across L and is π/2 ahead of I.
The amplitude of VL is vLm= imXL.
VC is voltage across C and is π/2 behind I.
The amplitude of VC is vCm= imXC.
Phasor-diagram Solution of Series LCR Circuit
The phasor relation of the voltages is
VR+VL+VC= V
Phasor-diagram Solution of Series LCR Circuit
The phasor relation of the voltages is
VR+VL+VC= V
VC and VL are always along the same line and
in opposite directions.
Phasor-diagram Solution of Series LCR Circuit
The phasor relation of the voltages is
VR+VL+VC= V
VC and VL are always along the same line and
in opposite directions.
The magnitude of (VC+VL) is |vCm - vLm|.
Phasor-diagram Solution of Series LCR Circuit
The phasor relation of the voltages is
VR+VL+VC= V
VC and VL are always along the same line and
in opposite directions.
The magnitude of (VC+VL) is |vCm - vLm|.
By pythagorean theorem,
vm2 = vRm2 + (vCm - vLm)2
Phasor-diagram Solution of Series LCR Circuit
The phasor relation of the voltages is
VR+VL+VC= V
VC and VL are always along the same line and
in opposite directions.
The magnitude of (VC+VL) is |vCm - vLm|.
By pythagorean theorem,
vm2 = vRm2 + (vCm - vLm)2
vm2 = [(imR)2 + (imXC - imXL)2 ]
Phasor-diagram Solution of Series LCR Circuit
The phasor relation of the voltages is
VR+VL+VC= V
VC and VL are always along the same line and
in opposite directions.
The magnitude of (VC+VL) is |vCm - vLm|.
By pythagorean theorem,
vm2 = vRm2 + (vCm - vLm)2
vm2 = [(imR)2 + (imXC - imXL)2 ]
vm2 = im2 [(R)2 + (XC - XL)2 ]
Phasor-diagram Solution of Series LCR Circuit
is called impedance (Z).
Phasor-diagram Solution of Series LCR Circuit
is called impedance (Z).
Phasor-diagram Solution of Series LCR Circuit
Phasor-diagram Solution of Series LCR Circuit
is called impedance (Z).
So,
Phasor-diagram Solution of Series LCR Circuit
is called impedance (Z).
So,
The phase angle ϕ is the angle between VR and V.
Phasor-diagram Solution of Series LCR Circuit
is called impedance (Z).
So,
The phase angle ϕ is the angle between VR and V.
Phasor-diagram Solution of Series LCR Circuit
is called impedance (Z).
So,
The phase angle ϕ is the angle between VR and V.
Phasor-diagram Solution of Series LCR Circuit
Impedance diagram
Right-triangle Impedance diagram with Z as its hypotenuse.
Phasor-diagram Solution of Series LCR Circuit
Impedance diagram
Right-triangle Impedance diagram with Z as its hypotenuse.
If XC > XL, ϕ is positive and the circuit is capacitive in nature.
Phasor-diagram Solution of Series LCR Circuit
Impedance diagram
Right-triangle Impedance diagram with Z as its hypotenuse.
If XC > XL, ϕ is positive and the circuit is capacitive in nature.
If XC< XL, ϕ is negative and the circuit is inductive in nature.
Phasor-diagram Solution of Series LCR Circuit
Impedance diagram
Right-triangle Impedance diagram with Z as its hypotenuse.
If XC > XL, ϕ is positive and the circuit is capacitive in nature.
If XC< XL, ϕ is negative and the circuit is inductive in nature.
If XC= XL, ϕ is zero and the circuit is purely resistive in nature.
Phasor-diagram Solution of Series LCR Circuit
Phasor diagram and graph for Series LCR Circuit (for XC > XL)
Current I leads voltage V by an angle ϕ for any arbitrary value of t.
Phasor-diagram Solution of Series LCR Circuit
Disadvantages of phasor diagram solution:
Phasor-diagram Solution of Series LCR Circuit
Disadvantages of phasor diagram solution:
Phasor-diagram Solution of Series LCR Circuit
Disadvantages of phasor diagram solution:
The solution so obtained is called the steady-state solution.
12P07.4
CV2
Analytical Solution of Series LCR Circuit
Analytical Solution of Series LCR Circuit
Analytical Solution of Series LCR Circuit
As, i = dq / dt and therefore di / dt = d2q / dt2
Analytical Solution of Series LCR Circuit
As, i = dq / dt and therefore di / dt = d2q / dt2
Let us assume a solution of this equation is
Analytical Solution of Series LCR Circuit
As, i = dq / dt and therefore di / dt = d2q / dt2
Let us assume a solution of this equation is
Analytical Solution of Series LCR Circuit
Substituting these values in voltage eq.
qm ω cos (ωt + θ) R − qm ω2 sin (ωt + θ) L + (qm / C) sin (ωt + θ) = vm sin ωt
Analytical Solution of Series LCR Circuit
Substituting these values in voltage eq.
qm ω cos (ωt + θ) R − qm ω2 sin (ωt + θ) L + (qm / C) sin (ωt + θ) = vm sin ωt
Using the relations XC= 1/ωC and XL = ωL, the voltage eq.
qm ω[R cos (ωt + θ) + (XC−XL) sin (ωt + θ)] = vm sinωt
Analytical Solution of Series LCR Circuit
Substituting these values in voltage eq.
qm ω cos (ωt + θ) R − qm ω2 sin (ωt + θ) L + (qm / C) sin (ωt + θ) = vm sin ωt
Using the relations XC= 1/ωC and XL = ωL, the voltage eq.
qm ω[R cos (ωt + θ) + (XC−XL) sin (ωt + θ)] = vm sinωt
Multiplying and dividing this eq. by
Analytical Solution of Series LCR Circuit
Substituting these values in voltage eq.
qm ω cos (ωt + θ) R − qm ω2 sin (ωt + θ) L + (qm / C) sin (ωt + θ) = vm sin ωt
Using the relations XC= 1/ωC and XL = ωL, the voltage eq.
qm ω[R cos (ωt + θ) + (XC−XL) sin (ωt + θ)] = vm sinωt
Multiplying and dividing this eq. by
Analytical Solution of Series LCR Circuit
Substituting these values in voltage eq.
qm ω cos (ωt + θ) R − qm ω2 sin (ωt + θ) L + (qm / C) sin (ωt + θ) = vm sin ωt
Using the relations XC= 1/ωC and XL = ωL, the voltage eq.
qm ω[R cos (ωt + θ) + (XC−XL) sin (ωt + θ)] = vm sinωt
Multiplying and dividing this eq. by
By using the relations and
Analytical Solution of Series LCR Circuit
Substituting these values in voltage eq.
qm ω cos (ωt + θ) R − qm ω2 sin (ωt + θ) L + (qm / C) sin (ωt + θ) = vm sin ωt
Using the relations XC= 1/ωC and XL = ωL, the voltage eq.
qm ω[R cos (ωt + θ) + (XC−XL) sin (ωt + θ)] = vm sinωt
Multiplying and dividing this eq. by
By using the relations and
qm ωZcos (ωt + θ − ϕ) = vm sinωt
Analytical Solution of Series LCR Circuit
Substituting these values in voltage eq.
qm ω cos (ωt + θ) R − qm ω2 sin (ωt + θ) L + (qm / C) sin (ωt + θ) = vm sin ωt
Using the relations XC= 1/ωC and XL = ωL, the voltage eq.
qm ω[R cos (ωt + θ) + (XC−XL) sin (ωt + θ)] = vm sinωt
Multiplying and dividing this eq. by
By using the relations and
qm ωZcos (ωt + θ − ϕ) = vm sinωt
Comparing the two sides of this equation
Analytical Solution of Series LCR Circuit
Analytical Solution of Series LCR Circuit
Analytical Solution of Series LCR Circuit
Analytical Solution of Series LCR Circuit
Analytical Solution of Series LCR Circuit
PSV 9
Q. A series circuit having capacitance 20 μF and inductance 30 mH is connected to a 220 V, 50 Hz source.
Find the impedance of the circuit.
Sol. The impedance
Q. A series circuit having capacitance 20 μF and inductance 30 mH is connected to a 220 V, 50 Hz source.
Find the impedance of the circuit.
Sol. The impedance
Since R = 0 ,
Q. A series circuit having capacitance 20 μF and inductance 30 mH is connected to a 220 V, 50 Hz source.
Find the impedance of the circuit.
Sol. The impedance
Since R = 0 ,
XC= (1 / 2 π f C)
= (1 / 2π × 50 × 20 × 10-6 )
= 1570.8 Ω
Q. A series circuit having capacitance 20 μF and inductance 30 mH is connected to a 220 V, 50 Hz source.
Find the impedance of the circuit.
Sol. The impedance
Since R = 0 ,
XC= (1 / 2 π f C)
= (1 / 2π × 50 × 20 × 10-6 )
= 1570.8 Ω
and XL = (2 π f L)
= ( 2 π × 50 × 30 × 10-3 )
= 9.425 Ω
Q. A series circuit having capacitance 20 μF and inductance 30 mH is connected to a 220 V, 50 Hz source.
Find the impedance of the circuit.
Sol. The impedance
Since R = 0 ,
XC= (1 / 2 π f C)
= (1 / 2π × 50 × 20 × 10-6 )
= 1570.8 Ω
and XL = (2 π f L)
= ( 2 π × 50 × 30 × 10-3 )
= 9.425 Ω
Q. A series circuit having capacitance 20 μF and inductance 30 mH is connected to a 220 V, 50 Hz source.
Find the impedance of the circuit.
12P07.4
CV3
Resonance
Resonance
Resonance occurs when,
The frequency of the energy source ≈ the natural frequency of the system
Resonance
Resonance occurs when,
The frequency of the energy source ≈ the natural frequency of the system
For a series RLC circuit
Resonance
Resonance occurs when,
The frequency of the energy source ≈ the natural frequency of the system
For a series RLC circuit
At ω = ω0, the impedance is minimum
Resonance
Resonance occurs when,
The frequency of the energy source ≈ the natural frequency of the system
For a series RLC circuit
At ω = ω0, the impedance is minimum
and or
or
Resonance
Resonance occurs when,
The frequency of the energy source ≈ the natural frequency of the system
For a series RLC circuit
At ω = ω0, the impedance is minimum
and or
or
At ω0 the current amplitude is maximum, im = vm / R
Resonance
Variation of im with ω in a Series RLC circuit (with L = 1.0 mH, C = 1.0 nF )
The resonant frequency ω0= {1 / √(LC)} = 1 × 106 rad/s
Resonance
Variation of im with ω in a Series RLC circuit (with L = 1.0 mH, C = 1.0 nF )
The resonant frequency ω0= {1 / √(LC)} = 1 × 106 rad/s
If the source applied is vm = 100 V
(i) For R = 100 Ω
Resonance
Variation of im with ω in a Series RLC circuit (with L = 1.0 mH, C = 1.0 nF )
The resonant frequency ω0= {1 / √(LC)} = 1 × 106 rad/s
If the source applied is vm = 100 V
(i) For R = 100 Ω
im = vm / R
im = 100/ 100 = 1 A
Resonance
Variation of im with ω in a Series RLC circuit (with L = 1.0 mH, C = 1.0 nF )
The resonant frequency ω0= {1 / √(LC)} = 1 × 106 rad/s
If the source applied is vm = 100 V
(i) For R = 100 Ω
im = vm / R
im = 100/ 100 = 1 A
(ii) For R = 200 Ω
Resonance
Variation of im with ω in a Series RLC circuit (with L = 1.0 mH, C = 1.0 nF )
The resonant frequency ω0= {1 / √(LC)} = 1 × 106 rad/s
If the source applied is vm = 100 V
(i) For R = 100 Ω
im = vm / R
im = 100/ 100 = 1 A
(ii) For R = 200 Ω
im = vm / R
im = 100/ 200 = 0.5 A
Resonance
Resonant circuits have applications in the tuning mechanism of a radio or a TV set
Resonance
Resonance phenomenon is exhibited by a circuit only
if both L and C are present in the circuit
Resonance
Sharpness of Resonance
The amplitude of the current in the series LCR circuit is
Resonance
Sharpness of Resonance
The amplitude of the current in the series LCR circuit is
im is maximum when ω = ω0= {1 / √(LC)}
The maximum value of im is
im = (vm/ R)
max
Resonance
Sharpness of Resonance
The amplitude of the current in the series LCR circuit is
im is maximum when ω = ω0= {1 / √(LC)}
The maximum value of im is
im = (vm/ R)
For values of ω other than ω0, the amplitude
of the current is less than the maximum value.
max
Resonance
For im = ( immax / √2)
The power dissipated by the circuit becomes half.
Resonance
For im = ( immax / √2)
The power dissipated by the circuit becomes half.
There are such two values of ω.
ω1 = ω0 + ∆ω
ω2 = ω0 – ∆ω
Resonance
For im = ( immax / √2)
The power dissipated by the circuit becomes half.
There are such two values of ω.
ω1 = ω0 + ∆ω
ω2 = ω0 – ∆ω
ω1 – ω2 = 2∆ω is called the bandwidth.
Resonance
For im = ( immax / √2)
The power dissipated by the circuit becomes half.
There are such two values of ω.
ω1 = ω0 + ∆ω
ω2 = ω0 – ∆ω
ω1 – ω2 = 2∆ω is called the bandwidth.
(ω0 / 2∆ω) is regarded as sharpness of resonance.
Resonance
For im = ( immax / √2)
The power dissipated by the circuit becomes half.
There are such two values of ω.
ω1 = ω0 + ∆ω
ω2 = ω0 – ∆ω
ω1 – ω2 = 2∆ω is called the bandwidth.
(ω0 / 2∆ω) is regarded as sharpness of resonance.
The smaller the ∆ω,
The sharper or narrower is the resonance.
Resonance
To get an expression for ∆ω
at
Resonance
To get an expression for ∆ω
at
Resonance
To get an expression for ∆ω
at
or
Resonance
To get an expression for ∆ω
at
or
or
Resonance
To get an expression for ∆ω
at
or
or
Resonance
To get an expression for ∆ω
at
or
or
Resonance
To get an expression for ∆ω
at
or
or
Using ω02 = 1/ LC in the second term on
the left hand side
Resonance
To get an expression for ∆ω
at
or
or
Using ω02 = 1/ LC in the second term on
the left hand side
We can approximate
Resonance
To get an expression for ∆ω
at
or
or
as
since
Resonance
Therefore,
Resonance
Therefore,
or,
Resonance
Therefore,
or,
The sharpness of resonance
Resonance
Therefore,
or,
The sharpness of resonance
The ratio (ω0L/ R) is the quality factor of
the circuit.
So,
Resonance
Therefore,
or,
The sharpness of resonance
The ratio (ω0L/ R) is the quality factor of
the circuit.
So,
Larger Q smaller 2∆ω (bandwidth)
Smaller bandwidth sharper resonance
Resonance
Therefore,
or,
The sharpness of resonance
The ratio (ω0L/ R) is the quality factor of
the circuit.
So,
Larger Q smaller 2∆ω (bandwidth)
Smaller bandwidth sharper resonance
Using ω02 = 1/ LC
Q = 1/ ω0CR
Resonance
Therefore,
or,
The sharpness of resonance
The ratio (ω0L/ R) is the quality factor of
the circuit.
PSV 10
Q. A circuit of resistance 30 Ω and
inductance 20 mH is in resonance
with angular frequency of 5000 rad/s.
Find the quality factor and bandwidth.
Sol. The quality factor
Q = (ω0 L / R)
Q. A circuit of resistance 30 Ω and
inductance 20 mH is in resonance
with angular frequency of 5000 rad/s.
Find the quality factor and bandwidth.
Sol. The quality factor
Q = (ω0 L / R)
= ( 5000 Χ 20 Χ 10-3 / 30 )
= 3.33
Q. A circuit of resistance 30 Ω and
inductance 20 mH is in resonance
with angular frequency of 5000 rad/s.
Find the quality factor and bandwidth.
Sol. The quality factor
Q = (ω0 L / R)
= ( 5000 Χ 20 Χ 10-3 / 30 )
= 3.33
Band width
2∆ω = (ω0 / Q)
Q. A circuit of resistance 30 Ω and
inductance 20 mH is in resonance
with angular frequency of 5000 rad/s.
Find the quality factor and bandwidth.
Sol. The quality factor
Q = (ω0 L / R)
= ( 5000 Χ 20 Χ 10-3 / 30 )
= 3.33
Band width
2∆ω = (ω0 / Q)
= (5000 / 3.33)
= 1500 rad/s
Q. A circuit of resistance 30 Ω and
inductance 20 mH is in resonance
with angular frequency of 5000 rad/s.
Find the quality factor and bandwidth.
PSV 11
Q. A resistor of 200 Ω and a capacitor of
15 µF are connected in series to a
220 V, 50 Hz ac source.
Q. A resistor of 200 Ω and a capacitor of
15 µF are connected in series to a
220 V, 50 Hz ac source.
Q. A resistor of 200 Ω and a capacitor of
15 µF are connected in series to a
220 V, 50 Hz ac source.
Sol.
Q. A resistor of 200 Ω and a capacitor of
15 µF are connected in series to a
220 V, 50 Hz ac source.
Sol.
Q. A resistor of 200 Ω and a capacitor of
15 µF are connected in series to a
220 V, 50 Hz ac source.
Sol.
The current in the circuit
Q. A resistor of 200 Ω and a capacitor of
15 µF are connected in series to a
220 V, 50 Hz ac source.
= (0.755 A)(200 Ω) =151 V
= (0.755 A)(200 Ω) =151 V
VC = I XC
= (0.755 A)(212.3 Ω) =160.3 V
= (0.755 A)(200 Ω) =151 V
VC = I XC
= (0.755 A)(212.3 Ω) =160.3 V
which is more than source voltage.
= (0.755 A)(200 Ω) =151 V
VC = I XC
= (0.755 A)(212.3 Ω) =160.3 V
which is more than source voltage.
So, they cannot be added like ordinary numbers.
By Pythagorean theorem
Total voltage
= (0.755 A)(200 Ω) =151 V
VC = I XC
= (0.755 A)(212.3 Ω) =160.3 V
which is more than source voltage.
So, they cannot be added like ordinary numbers.
a current i = imsin(ωt + ϕ ) in the circuit.
where and
Z is called the impedance of the circuit.
frequency ω0. The amplitude of the current is maximum at resonant frequency .
resonance.
Summary
12P07.4 Series LCR Circuit
Reference Questions
NCERT : 7.3,7.6,7.10,7.11,7.13.7.14,7.15,7.16,7.17,7.21,7.22
Work Book : 1,4,10,11,12,16
12P07.5
Power,LC Oscillations and Transformer
Series LCR Circuit
Learning Objective
Power and Power Factor for an AC Circuit
LC Oscillations
Transformers
12P07.5
CV1
Power and Power Factor for an AC Circuit
Power and Power Factor for an AC Circuit
For a series LCR circuit
v = vm sin ωt and i = imsin(ωt + ϕ )
Power and Power Factor for an AC Circuit
For a series LCR circuit
v = vm sin ωt and i = imsin(ωt + ϕ )
The instantaneous power p supplied by the source
p = v i = (vmsin ωt) × [ imsin(ωt + ϕ )]
Power and Power Factor for an AC Circuit
For a series LCR circuit
v = vm sin ωt and i = imsin(ωt + ϕ )
The instantaneous power p supplied by the source
p = v i = (vmsin ωt) × [ imsin(ωt + ϕ )]
Power and Power Factor for an AC Circuit
For a series LCR circuit
v = vm sin ωt and i = imsin(ωt + ϕ )
The instantaneous power p supplied by the source
p = v i = (vmsin ωt) × [ imsin(ωt + ϕ )]
The average power
Power and Power Factor for an AC Circuit
For a series LCR circuit
v = vm sin ωt and i = imsin(ωt + ϕ )
The instantaneous power p supplied by the source
p = v i = (vmsin ωt) × [ imsin(ωt + ϕ )]
The average power
Power and Power Factor for an AC Circuit
For a series LCR circuit
v = vm sin ωt and i = imsin(ωt + ϕ )
The instantaneous power p supplied by the source
p = v i = (vmsin ωt) × [ imsin(ωt + ϕ )]
The average power
The quantity cos ϕ is called the power factor.
Power and Power Factor for an AC Circuit
ϕ = 0 or cos ϕ = 1.
There is maximum power dissipation.
Case (i)
Resistive circuit
Power and Power Factor for an AC Circuit
ϕ = π / 2 or cos ϕ = 0
Therefore, no power is dissipated even though a current is flowing in the circuit.
This current is sometimes referred to as wattless current.
Case (ii)
Purely inductive or purely capacitive circuit
Power and Power Factor for an AC Circuit
Power dissipated
where
So, ϕ may be non-zero in a R L or R C or LC R circuit.
Even in such cases, power is dissipated only in the resistor.
Case (iii)
Series LCR circuit
Power and Power Factor for an AC Circuit
At resonance XC – XL= 0 , and ϕ = 0.
Therefore, cos ϕ = 1 and P = I 2 Z = I 2 R.
That is, maximum power is dissipated in a circuit (through R) at resonance.
Case (iv)
Power dissipated at resonance in LCR circuit
PSV 12
Q. A sinusoidal voltage of RMS value 200 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 µF. Find
Q. A sinusoidal voltage of RMS value 200 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 µF. Find
Q. A sinusoidal voltage of RMS value 200 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 µF. Find
Sol. XL = 2π f L
Q. A sinusoidal voltage of RMS value 200 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 µF. Find
Sol. XL = 2π f L
XL = 2π × 50 × 25.48 × 10-3
XL = 8 Ω
Q. A sinusoidal voltage of RMS value 200 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 µF. Find
Sol. XL = 2π f L
XL = 2π × 50 × 25.48 × 10-3
XL = 8 Ω
XC = (1 / 2π f C)
Q. A sinusoidal voltage of RMS value 200 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 µF. Find
Sol. XL = 2π f L
XL = 2π × 50 × 25.48 × 10-3
XL = 8 Ω
XC = (1 / 2π f C)
XC = (1 / 2π × 50 × 796 × 10-6)
XC = 4 Ω
Q. A sinusoidal voltage of RMS value 200 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 µF. Find
Sol. XL = 2π f L
XL = 2π × 50 × 25.48 × 10-3
XL = 8 Ω
XC = (1 / 2π f C)
XC = (1 / 2π × 50 × 796 × 10-6)
XC = 4 Ω
Q. A sinusoidal voltage of RMS value 200 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 µF. Find
Sol. XL = 2π f L
XL = 2π × 50 × 25.48 × 10-3
XL = 8 Ω
XC = (1 / 2π f C)
XC = (1 / 2π × 50 × 796 × 10-6)
XC = 4 Ω
Q. A sinusoidal voltage of RMS value 200 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 µF. Find
Sol. XL = 2π f L
XL = 2π × 50 × 25.48 × 10-3
XL = 8 Ω
XC = (1 / 2π f C)
XC = (1 / 2π × 50 × 796 × 10-6)
XC = 4 Ω
(i) Power factor = cos ϕ
Q. A sinusoidal voltage of RMS value 200 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 µF. Find
Sol. XL = 2π f L
XL = 2π × 50 × 25.48 × 10-3
XL = 8 Ω
XC = (1 / 2π f C)
XC = (1 / 2π × 50 × 796 × 10-6)
XC = 4 Ω
(i) Power factor = cos ϕ
= cos (-53.10)
= 0.6
Sol.
Sol.
Sol.
I = (V / Z)
Sol.
I = (V / Z)
I = (200 / 5)
I = 40 A
Sol.
(ii) P = V I cos ϕ
I = (V / Z)
I = (200 / 5)
I = 40 A
Sol.
(ii) P = V I cos ϕ
= 200 × 40 × 0.6
= 4800 Watt
I = (V / Z)
I = (200 / 5)
I = 40 A
12P07.5
CV2
LC Oscillations
LC Oscillations
Definition
When a capacitor (initially charged) is connected to an inductor, the charge on the capacitor and the current in the circuit exhibit the phenomenon of electrical oscillations or LC oscillations.
LC Oscillations
A capacitor can store electrical energy
An inductor can store magnetic energy
LC Oscillations
Let a capacitor be charged qm (at t = 0)
and connected to an inductor as shown
LC Oscillations
At time t, charge is q and current is i in the circuit.
LC Oscillations
At time t, charge is q and current is i in the circuit.
Since (di / dt) is positive.
By Kirchhoff’s loop rule,
LC Oscillations
At time t, charge is q and current is i in the circuit.
Since (di / dt) is positive.
By Kirchhoff’s loop rule,
∵ ( as q decreases, i increases )
∴
LC Oscillations
At time t, charge is q and current is i in the circuit.
Since (di / dt) is positive.
By Kirchhoff’s loop rule,
∵ ( as q decreases, i increases )
∴
SHM equation
On comparing
LC Oscillations
The charge on the capacitor
q = qm cos(ω0t + ϕ )
LC Oscillations
The charge on the capacitor
q = qm cos(ω0t + ϕ )
where qm is the maximum value of q
and ϕ is a phase constant.
LC Oscillations
The charge on the capacitor
q = qm cos(ω0t + ϕ )
where qm is the maximum value of q
and ϕ is a phase constant.
At t = 0, q = qm
So, cos ϕ = 1 or ϕ = 0
LC Oscillations
The charge on the capacitor
q = qm cos(ω0t + ϕ )
where qm is the maximum value of q
and ϕ is a phase constant.
At t = 0, q = qm
So, cos ϕ = 1 or ϕ = 0
Therefore, q = qm cos(ω0t) and i = im sin(ω0t)
LC Oscillations
Oscillation Steps
At t = 0, and
LC Oscillations
Oscillation Steps
At 0 < t > T / 4 , q decreases and i increases
LC Oscillations
Oscillation Steps
At t = T / 4 , q = 0 and i = im
LC Oscillations
Oscillation Steps
At T / 4 < t > T / 2 , q increases and i decreases
LC Oscillations
Oscillation Steps
At t = T / 2 , q = qm and i = 0
LC Oscillations
Oscillation Steps
At T / 2 < t > 3T / 4 , q decreases and i increases
LC Oscillations
Oscillation Steps
At t = 3T / 4 , q = 0 and i = im
LC Oscillations
Oscillation Steps
At 3T / 4 < t > T , q increases and i decreases
LC Oscillations
Oscillation Steps
At t = T , q = qm and i = 0
LC Oscillations
Waveform Representation of q and i
LC Oscillations
Analogies between Mechanical and Electrical Quantities
Mechanical system | Electrical system |
Mass m | Inductance L |
Force constant k | Reciprocal capacitance (1 / C) |
Displacement x | Charge q |
Velocity v = (dx / dt) | Current i = (dq / dt) |
Mechanical energy | Electromagnetic energy |
| |
LC Oscillations
LC Oscillations is not realistic for two reasons
LC Oscillations
LC Oscillations is not realistic for two reasons
PSV 13
Q. In a LC circuit having capacitance 10 μF and inductance 30 mH, the maximum charge on capacitor is 200 μC. Find the magnetic energy stored in the inductor when the charge on the capacitor is 50 μC.
Sol.
Sol.
Initially when capacitor is fully charged
Sol.
Initially when capacitor is fully charged
When charge on capacitor is 50 μC
Sol.
Initially when capacitor is fully charged
When charge on capacitor is 50 μC
Sol.
Initially when capacitor is fully charged
When charge on capacitor is 50 μC
12P07.5
CV3
Transformers
Transformers
It changes one voltage level to other voltage level (higher or lower) without changing it ’s frequency
Transformers
It changes one voltage level to other voltage level (higher or lower) without changing it ’s frequency
It works on the principle of mutual induction
Transformers
It changes one voltage level to other voltage level (higher or lower) without changing it ’s frequency
It works on the principle of mutual induction
It consists of two sets of coils, insulated from each other
Transformers
Coils are wound on a soft-iron core, either one on top of the other as in Fig (a) or on separate limbs of the core as in Fig (b).
Transformers
Primary coil (input coil) has Np turns and the secondary coil (output coil) has Ns turns.
Transformers
Operation Method
Transformers
Operation Method
Transformers
Operation Method
Transformers
Operation Method
Transformers
The induced emf or voltage ɛs , in the secondary with Ns turns.
Transformers
The induced emf or voltage ɛs , in the secondary with Ns turns.
The alternating flux Φ also induces an emf, called back emf in the primary
Transformers
The induced emf or voltage ɛs , in the secondary with Ns turns.
The alternating flux Φ also induces an emf, called back emf in the primary
But ɛp = vp and ɛs = vs
Transformers
The induced emf or voltage ɛs , in the secondary with Ns turns.
The alternating flux Φ also induces an emf, called back emf in the primary
But ɛp = vp and ɛs = vs
Therefore and
Transformers
The induced emf or voltage ɛs , in the secondary with Ns turns.
The alternating flux Φ also induces an emf, called back emf in the primary
But ɛp = vp and ɛs = vs
Therefore and
So,
Transformers
If the transformer is assumed to be 100% efficient.
Input power = Output power
Transformers
If the transformer is assumed to be 100% efficient.
Input power = Output power
So,
Transformers
If the transformer is assumed to be 100% efficient.
Input power = Output power
So,
and
Transformers
If the transformer is assumed to be 100% efficient.
Input power = Output power
So,
and
If (Ns > Np), the voltage is stepped up (Vs > Vp).
This type of arrangement is called a step-up transformer.
Transformers
If the transformer is assumed to be 100% efficient.
Input power = Output power
So,
and
If (Ns > Np), the voltage is stepped up (Vs > Vp).
This type of arrangement is called a step-up transformer.
If (Ns < Np) , the voltage is stepped down (Vs < Vp) .
This type of arrangement is called a step-down transformer.
PSV 14
Q. A power transmission line feeds input power at 11000 V to a step-down transformer
with its primary windings having 4000 turns. what should be the number of turns in the
secondary in order to get output power at 220 V ?
Q. A power transmission line feeds input power at 11000 V to a step-down transformer
with its primary windings having 4000 turns. what should be the number of turns in the
secondary in order to get output power at 220 V ?
Sol. As we know
Q. A power transmission line feeds input power at 11000 V to a step-down transformer
with its primary windings having 4000 turns. what should be the number of turns in the
secondary in order to get output power at 220 V ?
Sol. As we know
Q. A power transmission line feeds input power at 11000 V to a step-down transformer
with its primary windings having 4000 turns. what should be the number of turns in the
secondary in order to get output power at 220 V ?
Sol. As we know
Q. A power transmission line feeds input power at 11000 V to a step-down transformer
with its primary windings having 4000 turns. what should be the number of turns in the
secondary in order to get output power at 220 V ?
Sol. As we know
Ns = 80
Transformers
Reasons for Energy Losses in Actual Transformers
Flux Leakage
There is always some flux leakage. All of the flux due to primary doesn’t pass through the secondary due to poor design of the core or the air gaps in the core.
Transformers
Reasons for Energy Losses in Actual Transformers
Flux Leakage
Resistance of
the windings
The wire used for the windings has some resistance and so, energy is lost due to heat produced in the wire (I2 R).
Transformers
Reasons for Energy Losses in Actual Transformers
Flux Leakage
Resistance of
the windings
Eddy currents
The alternating magnetic flux induces eddy currents in the iron core and causes heating. The effect is reduced by using a laminated core.
Transformers
Reasons for Energy Losses in Actual Transformers
Flux Leakage
Resistance of
the windings
Eddy currents
Hysteresis
The magnetisation of the core is repeatedly reversed by the alternating magnetic field. The resulting expenditure of energy in the core appears as heat
The average power loss over a complete cycle is given by P = V I cos Φ.
In such cases, current is referred to as a wattless current.
Summary
12P07.5 Power,LC Oscillations and Transformer
Reference Questions
NCERT : 7.7, 7.8, 7.9, 7.12, 7.18, 7.19, 7.20 ,7.23, 7.24, 7.25,
7.26
Work Book : 2, 3, 7, 15, 17, 18, 19