chemistry lecture

1. ELL100: INTRODUCTION TO
ELECTRICAL ENG.
Lecture 7
Course Instructors:
Anandarup Das, Rakesh Palani, Jayadeva,
Shouri Chatterjee, Ankur Gupta
IIT Delhi

2. AC power transmission
• Throughout the world, generation and transmission of electrical power
takes place in AC.
• AC power is generated at the generating station (thermal, hydro etc.) and
transmitted over hundreds of kilometers by stepping up the voltage
through a transformer. At the consumer end, the voltage is stepped down.
• Why AC transmission is more popular and not DC transmission?
• AC transmission at high voltage is economical because of lesser line drop and losses
in the transmission line.
• Because of circuit breakers, protection in AC is easier because current goes through
zero twice in a cycle. Thus fault clearance is easier in AC than DC.

3. Notations in AC
• Instantaneous voltage/current: v(t), i(t)
• RMS voltage: V or Vrms. For sinusoidal waveform, Vm=√2 V
• Instantaneous power: p(t)=v(t).i(t)
𝑇 0
• Average power: 𝑃 = 1
∫
𝑇
𝑝 𝑡 𝑑𝑡

4. Single phase AC network (resistive load)
• The voltage (blue) and the current
(green) waveforms are in phase.
• The instantaneous power (red) is
sinusoidal and its average value is
positive.
• Instantaneous power: 𝑝 𝑡
Where, Im=Vm/R.
= 𝑣 𝑡 . 𝑖 𝑡 = 𝑉𝑚 cos(𝜔𝑡) 𝐼𝑚 cos 𝜔𝑡
0
• Average power: 𝑃 = 1
∫
𝑇
𝑝 𝑡 𝑑𝑡 = 𝐼2𝑅 = 𝑉2
𝑇 𝑅

5. Single phase AC network (inductive load)
• The voltage (blue) waveform leads the
current (green) waveform by 900.
• The instantaneous power (red) is
sinusoidal and its average value is zero.
• Periodic exchange of energy takes place
between source and load.
• Instantaneous power: 𝑝 𝑡 = 𝑣 𝑡 . 𝑖 𝑡 = 𝑉𝑚 cos(𝜔𝑡) 𝐼𝑚 cos 𝜔𝑡 − 𝜋/2
Where, 𝐼𝑚 = 𝑉𝑚
𝜔𝐿
• Simplifying, 𝑝 𝑡 = 𝐼2𝜔𝐿 sin(2𝜔𝑡) = 𝐼2𝑋𝐿 sin(2𝜔𝑡)
𝑇 0
• Average power: 𝑃 = 1
∫
𝑇
𝑝 𝑡 𝑑𝑡 = 0

6. Single phase AC network (capacitive load)
• The voltage (blue) waveform lags the
current (green) waveform by 900.
• The instantaneous power (red) is
sinusoidal and its average value is zero.
• Periodic exchange of energy takes place
between source and load.
• Instantaneous power: 𝑝 𝑡 = 𝑣 𝑡 . 𝑖 𝑡 = 𝑉𝑚 cos(𝜔𝑡) 𝐼𝑚 cos 𝜔𝑡 + 𝜋/2
Where, 𝐼𝑚 = 𝑉𝑚𝜔𝐶
𝜔𝐶
Simplifying, 𝑝 𝑡 = − 𝐼2
sin(2𝜔𝑡) = 𝐼2𝑋𝐶 sin(2𝜔𝑡)
𝑇 0
• Average power: 𝑃 = 1
∫
𝑇
𝑝 𝑡 𝑑𝑡 = 0

7. Single phase AC network (RLC load)
• In general, the voltage (blue) waveform
leads/lags the current (green) waveform by
certain angle. The power (red) can have an
average value zero, positive or negative.
• Average power negative indicates the load
is delivering power back to source which
cannot continue for an indefinite period of
time.
• Instantaneous power: 𝑝 𝑡 = 𝑣 𝑡 . 𝑖 𝑡 = 𝑉𝑚 cos(𝜔𝑡 + 𝜃) 𝐼𝑚 cos 𝜔𝑡
Where, 𝐼𝑚 = 𝑉𝑚/𝑍
• Thus, 𝑝 𝑡 = 𝑉𝐼 cos 𝜃 + 𝑉𝐼 cos(2𝜔𝑡 + 𝜃)
• Average power: 𝑃 = 𝑉𝐼 cos 𝜃

8. Complex power
• The phasor diagram (Fig. a) shows an
inductive load where I lags V by an angle
θ.
• Divide the magnitude of each voltage
phasor by I and we obtain the impedance
triangle ( Fig. b).
• Multiply the magnitude of each voltage
phasor by I and we obtain the power
triangle (Fig. c).
(a) Phasor diagram
(b) Impedance triangle
(c) Power triangle

9. Complex power
• 𝑃𝐴 = 𝑉𝐼 cos 𝜃 + 𝑗𝑉𝐼 sin 𝜃 = 𝑉𝐼∠𝜃 = 𝑃 + 𝑗𝑃𝑋
• PA: apparent power, unit VA
• P : real power, unit W
• PX: reactive power, unit VAR
• Power factor = cos 𝜃= P/VI.
• When current lags the voltage, θ is positive and reactive power 𝑉𝐼 sin 𝜃 is
positive. In an inductive circuit, Px is positive i.e. inductive circuit absorbs
reactive power from the voltage source.
• When current leads the voltage, θ is negative and reactive power 𝑉𝐼 sin 𝜃
is negative. In a capacitive circuit, Px is negative i.e. capacitive circuit
supplies reactive power to the voltage source.

10. Example
(1)
(2)
(3)
(Note: Reactive power is positive)
(4)

11. Power factor correction
• A poor power factor is not desirable for the electric supply company. A poor
power factor indicates that for the same active power delivered, the
reactive power is large.
• Thus, more current needs to be delivered which increases losses in the
transmission.
• From the view point of electric supply company, loads should always ideally
operate at unity power factor.
• Reactive power minimization or power factor correction is therefore done
by consumers.

12. Example
• An industrial load consists of
20 kW of heating and
induction motors which take
100 kVA at 0.707 power factor
lag.
• Find out the total active and
reactive power drawn by the
industrial load. Find out the
plant power factor.
• 𝑃𝐴𝑚𝑜𝑡 = 100𝑋103∠45 = 𝑃𝑚𝑜𝑡 + 𝑗𝑃𝑋𝑚𝑜𝑡 = 70𝑋103 + 𝑗70𝑋103 VA
• 𝑃ℎ𝑒𝑎𝑡 = 20𝑋103∠0 W
• 𝑃𝐴𝑖𝑛𝑑 = 70 + 20 𝑋103 + 𝑗70𝑋103 = 114𝑋103∠37.8 VA
• Plant power factor = cos (37.8)= 0.79 lag.

13. Example
• If the power factor needs to be
improved to 0.95 lag, then what
should be installed? What is its
power rating?
• Capacitors can improve the power factor by supplying reactive power.
• 𝑃𝑋𝑖𝑛𝑑 = 𝑗𝑃𝑖𝑛𝑑 tan 𝜃 = 𝑗90𝑋103 tan ( cos−1 0.95) = 𝑗29.6𝑋103 VAR
• 𝑃𝑋𝑐𝑎𝑝 = −𝑗 70𝑋103 − 29.6𝑋103 = −𝑗40.4𝑋103 VAR
• Note that, 𝑃𝐴𝑖𝑛𝑑 = 90𝑋103 + 𝑗29.6𝑋103 = 98.6𝑋103∠18.2 VA is now
less than the previous case indicating a reduction of current drawn.

14. Three phase AC network
• Generation and transmission of power more efficient in three phase
network. In single phase, two conductors are needed; while in 3 phase,
three conductors can transmit 3 times the power.
• Power in a three-phase circuit is constant rather than pulsating.
• Three-phase generators are used for generation of electricity where a
rotating magnetic field is produced. Three phase motors start and run
better than single-phase.

15. Three phase AC circuit
• A 3 phase AC voltage source is
represented by three balanced
voltages. The voltages are
equal in magnitude, have same
frequency and 1200 apart in
time.
• In phasor representation, three
balanced phasors are 1200
apart and rotating at a speed
ω.
• 𝑉𝑎= 𝑉∠00, 𝑉𝑏 = 𝑉∠ −1200,
𝑉𝑐= 𝑉∠ − 2400

16. Three phase AC circuit – star/delta connection
• 3 phase AC voltage sources can
be connected in star or delta
connection.
• Loads can also be connected in
star or delta connection.
• Star connection is sometimes
called wye connection.
Star connected source connected to delta connected load
Delta connected source connected to delta connected load

17. Delta connections
• The following relationship holds for delta
connection, obtained by applying KVL and
KCL.
• The line voltages are equal to phase voltage but line currents are not equal to phase
currents.
• In general,

18. Star connection
• In star connected load, line
current is same as phase current.
• Line voltages are not same as
phase voltages.

19. Example
By symmetry,
The source is balanced with line to line voltage=220V.

20. Power in 3 phase circuit
• The total power in a balanced three-phase load is the sum of three equal
phase powers.
• 𝑃𝑡𝑜𝑡𝑎𝑙 = 3 𝑃𝑝ℎ𝑎𝑠𝑒 = 3 𝑉𝑝ℎ𝐼𝑝ℎ cos 𝜃𝑝ℎ
3
• In delta connected load,
• 𝑃𝑑𝑒𝑙𝑡𝑎 = 3 𝑉𝑝ℎ𝐼𝑝ℎ cos 𝜃𝑝ℎ = 3 𝑉𝑙𝑖𝑛𝑒
𝐼𝑙𝑖𝑛𝑒
cos 𝜃𝑝ℎ = 3 𝑉𝑙𝑖𝑛𝑒 𝐼𝑙𝑖𝑛𝑒cos 𝜃𝑝ℎ
3
• In star connected load,
• 𝑃𝑠𝑡𝑎𝑟 = 3 𝑉𝑝ℎ𝐼𝑝ℎ cos 𝜃𝑝ℎ = 3 𝑉𝑙𝑖𝑛𝑒
𝐼𝑙𝑖𝑛𝑒cos 𝜃𝑝ℎ = 3 𝑉𝑙𝑖𝑛𝑒 𝐼𝑙𝑖𝑛𝑒 cos 𝜃𝑝
ℎ

21. Example
Per phase power is,
The source is balanced with line to line voltage=220V.
3 phase power is,