Sinusoidal Alternating voltage & current :
CBSE GRADE 12
If the waveform of alternating voltage is a sine wave, then it is
PHYSICS STUDY MATERIAL
known as sinusoidal alternating voltage
CHAP 07 ALTERNATING CURRENT
V = Vo sin ѡt
Syllabus ( 2024 – 2025 ) : Alternating currents, peak and RMS
or in terms of EMF it is represented as E = Eo sin ѡt
value of alternating current/voltage; reactance and impedance; LCR
series circuit (phasors only), resonance, power in AC circuits, power When sinusoidal alternating voltage is applied to a closed
factor, wattless current. AC generator, Transformer circuit, the resulting alternating current is also sinusoidal in
nature
Alternating Voltage & Alternating Current :
I = Io sin ѡt
A voltage that changes its polarity at regular intervals of time is
MEAN VALUE or AVERAGE VALUE OF CURRENT
called as Alternating Voltage and this gives rise to a current that
The mean value of AC for a full cycle is zero. So the mean value is
changes its direction at regular intervals of time is called as
calculated only for half cycle
Alternating Current.
It is defined as the value of AC which would send same amount
It is represented as I = Io sin ѡt (or) I = Io cos ѡt of charge through a circuit in half-cycle that is sent by steady
I = Instantaneous value of the current current in the same time ( T/2 )
Io = Maximum or Peak value of the current
Ѡ = Angular frequency ( ѡ = 2πf )
Alternating voltage can be easily converted from one
value to another by a transformer.
Alternating current energy can be transmitted to long
distance without much energy losses.
ROOT MEAN SQUARE ( RMS ) VALUE OF CURRENT
It is defined as the value of AC over a complete cycle which would
generate same amount of heat in a given resistors that is generated
by steady current in the same resistor and in the same time during
a complete cycle.
“ Mean value of AC is 63% of peak value of AC over any half-
cycle. In a complete cycle of AC, the mean value of AC will be
zero ”
2. A light bulb is rated 150W for 220V ac supply of 60Hz.
Calculate (i) the resistance of the bulb
(ii) the rms current through the bulb.
[ Ans : R = 322.7 Ω , Irms = 0.68 A ]
Phasor and Phasor diagram : A sinusoidal alternating voltage
(or current) can be represented by a vector which rotates about
the origin in anti-clockwise direction at a constant angular
velocity ω. Such a rotating vector is called a phasor.
The diagram which shows various phasors and their phase
relations is called phasor diagram.
“ The 70.7% of peak value of current gives effective
Or RMS value of AC ”
AC CIRCUIT WITH PURE RESISTOR ONLY
Class work Numerical :
1. The equation for an alternating current is given by
i = 77 sin 314 t. Find the peak current, frequency, time period
and instantaneous value of current at t = 2 ms.
E
Consider a circuit containing a pure resistor of resistance R AC CIRCUIT WITH PURE INDUCTOR ONLY
connected across an alternating voltage source. The instantaneous
value of the alternating EMF is given by
E = Eo sin ѡt -----(1)
An alternating current I flowing in the circuit due to this voltage,
develops a potential drop across R
E = IR -----(2) E
Consider a circuit containing a pure inductor of inductance L
Combining both the equations we get connected across an alternating voltage source . The
Eo sin ѡt = IR instantaneous value of the alternating voltage is given by
𝐄𝐨 𝐬𝐢𝐧 ѡ𝐭 E = Eo sin ѡt -----(1)
I= = Io sin ѡt -----(3)
𝑹
The alternating current flowing through the inductor induces a
From equ(1) &(3), it is clear the current and voltage are in self-induced emf or back emf in the circuit.
same phase ( Φ = 00 ) 𝒅𝑰
E=-L ----- (2)
𝒅𝒕
To maintain current in the circuit the applied voltage must be
equal and opposite to the induced voltage
AC CIRCUIT WITH PURE CAPACITOR ONLY
----- (3)
From equ(1) &(3), it is clear the current lags the voltage by
900 E
Consider a circuit containing a pure capacitor of Capacitance
C connected across an alternating voltage source . The
instantaneous value of the alternating voltage is given by
E = Eo sin ѡt -----(1)
Let q be the instantaneous charge on the capacitor. The emf
across the capacitor at that instant is
V = q/C V=E
E = q/C -----(2)
INDUCTIVE REACTANCE ( XL ) Combining both the equations we get
ѡL refers to the resistance offered by an inductor known as
Inductive Reactance. Its SI unit is Ω
XL = ѡL = 2πfL
For a DC circuit f = 0, so XL becomes 0
So a pure inductor circuit cannot reduce DC
An inductor blocks AC but it allows DC ----- (3)
From equ(1) &(3), it is clear the current leads the voltage by
900
Class work Numerical :
3. A 44 mH inductor is connected to 220 V, 50 Hz AC supply.
Determine the rms value of the current in the circuit.
( Ans : Irms = 15.92 A )
CAPACITIVE REACTANCE ( XC ) 4. A 15 μF capacitor is connected to a 220V, 50Hz source. Find
1/ѡC refers to the resistance offered by an capacitor known as the capacitive reactance and the current (rms and peak) in
Capacitive Reactance. Its SI unit is Ω the circuit. If the frequency is doubled, when happens to the
Xc = 1/ѡC = 1/2πfC capacitive reactance and the current?
SERIES RLC CIRCUIT
AC circuit containing a resistor, an inductor and a
capacitor in series
Consider a circuit containing a resistor of resistance R, an
inductor of inductance L and a capacitor of capacitance C
connected across an alternating voltage source .
For a DC circuit f = 0, so XC becomes infinite
A Capacitor will block DC but it allows AC
REMEMBER THE FORMULA
The instantaneous value of the alternating current is given by
I = Io sin ѡt
Direction of the resultant :
EO Special Case :
If XL = XC, Φ is zero. Therefore current and voltage are in the
same phase and the circuit is resistive.
If XL > XC, Φ is also positive. It means that the applied
voltage leads the current . The circuit is inductive.
If XL < XC, Φ is also negative. It means that the applied
The maximum voltage across resistor VR = IOR voltage lags the current . The circuit is capacitive.
The maximum voltage across inductor VL = IOXL
RESONANCE IN RLC CIRCUIT
The maximum voltage across capacitor VC = IOXC
When the frequency of the applied alternating source is equal to
Solving the above phasor diagram using parallelogram law of the natural frequency of the RLC circuit, the current in the circuit
vectors ( ODEA ) reaches its maximum value. Then the circuit is said to be in
electrical resonance. The frequency at which resonance takes place
is called resonant frequency ( ѡr )
Z is called impedance of the circuit which refers to the effective
opposition to the current by the series RLC circuit.
Applications of series RLC resonant circuit :
RLC circuits have many applications like filter circuits,
oscillators, voltage multipliers etc. An important use of series
RLC resonant circuits is in the tuning circuits of radio and TV
systems. The signals from many broadcasting stations at
different frequencies are available in the air.
Note : The phenomenon of electrical resonance is possible when
the circuit contains both L and C. Only then the voltage across
L and C cancel one another when VL and VC are 180° out of
phase and the circuit becomes purely resistive. This implies
that resonance will not occur in RL and RC circuits.
The maximum current at series resonance is limited by the
resistance of the circuit. For smaller resistance, larger current with
sharper curve is obtained and vice versa.
Class work Numerical :
5. Three electrical circuits having AC sources of variable frequency
are shown in the figures. Initially, the current flowing in each of
these is same. If the frequency of the applied AC source is increased,
how will the current flowing in these circuits be affected? Give the
reason for your answer.
P = Vo sin ѡt ( Io sin ( ѡt + Φ ))
P = Vo Io sin ѡt (sin ( ѡt + Φ ))
P = Vo Io sin ѡt (sin ѡt cos Φ + cos ѡt sin Φ )
P = Vo Io (sin 2 ѡt cos Φ + sin ѡt cos ѡt sin Φ )
6. The figure shows a series LCR circuit with L = 10.0H, C=
Here the average of sin 2 ѡt over a cycle is 1/2 and sin ѡt
40μF, R = 60Ω connected to a variable frequency 240V source,
calculate (i) the angular frequency of the source which drives the cos ѡt is zero. Substituting these values, we obtain average
circuit at resonance, (ii) the current at the resonating frequency, (iii) power over a cycle.
the rms potential drop across the inductor at resonance.
where VRMS IRMS is called apparent power and cos Φ is power
7. A 500 μH inductor, 80/π2 pF capacitor and a 628 Ω resistor are factor. The average power of an AC circuit is also known as the
connected to form a series RLC circuit. Calculate the resonant true power of the circuit.
frequency
Special Case :
POWER IN AC CIRCUITS For a purely resistive circuit, Φ = 0 then cos Φ = 1
Power of a circuit is defined as the rate of consumption of electric Pavg = VRMS IRMS
energy in that circuit. It is given by the product of the voltage and
For a purely inductive / Capacitive circuit, Φ = π/2 0then cos
current.
Φ=0
The alternating voltage and alternating current in the series Pavg = 0
inductive RLC circuit at an instant are given by For series RLC circuit, the phase angle Φ
V = Vo sin ѡt
I = Io sin ( ѡt + Φ )
Power P = VI Pavg = VRMS IRMS cos Φ
For series RLC circuit at resonance, Φ = 0 and cos Φ =1 Power factor = True Power / Apparent Power
Pavg = VRMS IRMS
Wattless current : The current in an AC circuit is said to be
wattless current if the power consumed by it is zero. This wattless
current occurs in a purely inductive or capacitive circuit.
Special Case :
For a purely resistive circuit, Φ = 0 then cos Φ = 1 so the
Power factor = 1
For a purely inductive / Capacitive circuit, Φ = π/2 then
(i) The component of current IRMS cos Φ cos Φ = 0 so the Power factor = 0
which is in phase with the voltage is called active component. The Power factor lies between 0 and 1 for a circuit having R, L and
power consumed by this current= VRMS IRMS cos Φ So that it is also C in varying proportions.
known as ‘Wattful’ current.
Class work Numerical :
(ii) The other component IRMS sin Φ which has a phase angle of 8. An inductor 200mH, capacitor 500μF, resistor 10Ω are
π/2 with the voltage is called reactive component. The power connected in series with a 100V, variable frequency a.c. source.
consumed is zero. Hence it is also known as ‘Wattless’ current. Calculate the (i) frequency at which the power factor of the
circuit is unity. (ii) current amplitude at this frequency
POWER FACTOR
The power factor of a circuit is defined in one of the following 9. A capacitor C, a variable resistor R and a bulb B are connected
ways: in series to the AC mains in the circuit as shown. The bulb glows
with some brightness. How will the glow of the bulb change if (i)
Power factor = cos Φ = cosine of the angle of lead or lag a dielectric slab is introduce between the plates of the capacitor
keeping resistance R to be the same (ii) the resistance R is
Power factor = R/Z = Resistance / Impedance
increased keeping the same capacitance?
capacitor C, in order to make the power factor of the circuit
unity.
13. (i) When an AC source is connected to an ideal inductor show
that the average power supplied by the source over a complete
cycle is zero.
10. A series L-C-R circuit is connected to an AC source (200 V, 50 Hz). (ii) A lamp is connected in series with an inductor and an AC
The voltages across the resistor, capacitor and inductor are source. What happens to the brightness of the lamp when the key
respectively 200 V, 250 V and 250 V. (i) The algebraic sum of the is plugged in and an iron rod is inserted inside the inductor?
voltages across the three elements is greater than the voltage of Explain.
the source. How is this paradox resolved? (ii) Given the value of
the resistance of R is 40 Q, calculate the current in the circuit.
11. A resistor 'R' and an element 'X' are connected in series to an AC
source of voltage. The voltage is found to lead the current in phase
by π/4. If 'X is replaced by another element 'Y, the voltage lags
behind the current by π/4. 14. A device X is connected to an AC source, V = Vo sin ѡt. The
(i) Identify elements X and Y
variation of voltage, current and power in one cycle is shown in the
(ii) When both 'X and 'Y are connected in series with 'R' to the
following graph.
same source, will the power dissipated in the circuit be
maximum or minimum? Justify
12. Find the value of the phase difference between the current and
the voltage in the series L-C-R circuit shown below.
(i) Identify the device X.
(ii) Which of the curves A, Band C represent the voltage, current and
(i) Which one leads in phase: current or voltage? the power consumed in the circuit? Justify
(ii) Without making any other change, find the value of the (iii) How does its impedance vary with frequency of the AC source?
additional capacitor C', to be connected in parallel with the Show graphically.
(iv) Obtain an expression for the current in the circuit and its
phase relation with AC voltage
15. The graphs shown here depict the variation of current with
ang frequency for two different series LCR Observe the graphs
carefully. (i) State the relation b/w L and C values of the two circuits
when the current in the two circuits is maximum. (ii) Indicate which
power factor is higher
