Upload
lamdang
View
217
Download
2
Embed Size (px)
Citation preview
Alternating Current (ac) Batteries are a source of steady or direct voltage. Current in a circuit powered by a battery is also steady and is called direct current (dc)
t
V V0 Voltage (& current) are
constant
Direct voltage (current) dc
Nearly all the electricity we use is in the form of alternating current (& voltage) termed ac
Alternating Current (ac):
V
-V0
0 t
+V0
0
-V0 •generally sinusoidal. •current also alternates
Voltage periodic: • Magnitude and direction changes
Alternating Current (ac)
0 sinV V tω=V
-V0
0 t
+V0
0
-V0
2 2 /f Tω π π= =Angular frequency
I
0 t
+I0
0
-I0
0 sinI I tω=
Time averaged value of sine function over one or more cycles is zero
Average values of the current and voltage is zero
R
0 sinV V tω=
0 sinI I tω=
Alternating Current (ac)
-
R
+
V
dc
ac: Average value of the current is zero
However the power consumed in an ac circuit is not zero because dissipation of energy in the resistance does not depend on the direction of the current. ( Power = R I2 > 0 )
I
0 t
+I0
0
-I0
Electrical power consumed by any resistive component in a circuit is P =I2R
0 sinI I tω=
ac
2I20I20
12
I
t 0
( ) ( )220
12ave
I I=
( )2 210 02
12rms aveI I I I= = =
2 2 20 sinI I tω=
rms effI I=
Alternating Current (ac)
always positive 2I
Average value of 2I
Root mean square (rms) value
Similarly ( )012rmsV V=
rms effV V=
Square root of the average value of the square of the current
0 sinI I tω=
Alternating Current (ac)
0 sinI I tω=I
0 t
+I0
0
-I0
Ieff
T
ac voltage and current are always characterized by their effective (or root mean square ) values:
where Vo, and Io are the amplitudes or maximum values of the voltage and the current.
T is the period (= 1/f)
I0 is the maximum current value or current amplitude
Average current value is zero.
and √2 Vo Veff = Ieff =
√2 Io
Alternating Current
An alternating current with a maximum value of 3 amps will produce the same heating effect in a resistance as a direct current of (3/√2)amps
ac mains varies at a rate of 50 times per second– frequency of 50 cycles per second
So periodic time (time for one cycle) T= (1/50)s = 20x10-3 s = 20 millisecs
SI unit of frequency named after German Physicist Heinrich Hertz 1857-1894.
ac mains---frequency 50 Hertz ----50Hz
I
0 t
+I0
0
-I0
Ieff
T
and √2 Vo Veff = Ieff =
√2 Io
Alternating Current I
0 t
+I0
0
-I0
Ieff
T
and √2 Vo Veff = Ieff =
√2 Io
Meters that measure ac current and voltage Calibrated to measure: effective (or rms ) current and voltage Ieff and Veff
ac power Pave = IeffVeff
Power in an ac circuit
since V0=I0R
since I0 = V0/R
( IoVo) 2 Pave =
(Io)2R 2 Pave =
(Vo)2 2R
Pave =
Pave = (I0/√2) (V0/√2)
Power in a dc circuit 2P I R=P IV= 2VPR
=
ESB provides electricity • voltage of 220V •frequency of 50Hz.
The period of the oscillations is: T = 1/50Hz = 0.02s = 20 milliseconds.
220V is the effective or rms value of the voltage: Veff = 220V
amplitude V0 is: V0 = Veff *√2 = 311V
The effective current going through a 100W light bulb is:
Ieff = Pave/Veff = (100/220) A = 0.45A
Alternating Current
Mains voltage swings from +311V to -311V 50 times every second giving an effective voltage of 220V.
Nearly all the electricity we use is in the form of alternating current (& voltage ) termed ac.
Alternating Current
Why?
1. ac voltage (compared with dc voltage) can be increased or decreased easily (with a transformer) 2. High voltage can be transmitted more efficiently
Transmission of electrical energy from generating stations to homes and businesses.
Transmission more efficient at high voltages
Transmission lines (cables) 100’s km long Therefore resistance of cables is significant
energy loss is reduced by reducing current
2ave effP I R=
Alternating Current
Want to deliver a power P to the substation where the transmission line arrives
Power loss in transmission cable of resistance R :
To minimise loss need to increase voltage for a fixed power P to be delivered. Typically 100-700 kV
Step down transformer used to convert the voltage to 220V for use in home or workplace.
2loss effP RI=
2
2loss eff
eff
PP RI RV
= =
Example
A maximum ac voltage of 240 V at a frequency 50 Hz is connected across a 530 Ω resistor. Find (a) rms voltage. (b) rms current. (c) the average power and (d) the maximum power dissipated in the resistor. R
0 sinV V tω=
0 sinI I tω=(a) 0 2rmsV V= ×
0 240 1702 2rms
V VV V= = ≈
(b) 170 0.32530
rmsrms
V VI AR
= = ≈Ω
(c) ( )22 0.32 530 54.3ave rmsP I R A Watts= × = × Ω =
(d) 2 2 2
maxmax
1702 2 109530
rmsV VP WattsR R
= = = =
If the frequency is doubled, does the average power dissipated in the resistor increase, decrease or stay the same.
Electrical Power Example
A 100W bulb operates at a voltage of 220V. Determine the resistance of its filament.
Pave = V2/R or R =V2/Pave
R = (220)2/100 = 484Ω
Determine the rms and peak currents through the bulb
Pave = IrmsVrms
100 0.455220
averms
rms
PI ampsV
= = =
( ) ( )0 2 2 0.455 0.643rmsI I amps amps= = =
250 kV high voltage transmission cable 10km long, resistance 280 Ω supplying 100 MW of power. Bird’s feet are 7 cm apart.
Bird sitting on high voltage transmission cable. Why is it not electrocuted?
Example
Current in the cable P IV=100 400
25P MWattsI ampsV kV
= = =
Resistance of 7 cm length 2
43
7 10 280 19.6 1010 10
mRm
−−×
= × Ω = × Ω×
Voltage between feet 4400 19.6 10 0.784V IR V−= = × × =
Assume resistance through body = 8 kΩ
Current through bird = 60.784 98 108
V ampsk
−= ×Ω
Current through bird = 250 31.258
kV ampsk
=Ω
Electrical safety Hazards: Thermal & Electrical shock
Heat produced faster than it can be dissipated
R3 R2
I2 I3 I4 I5 I6
e.g. As more resistances (appliances) are added in parallel current increases
Thermal
Normally thermal energy produced in wires is negligible unless current becomes very large
Power dissipated in the feed cable P = I2Rw Power proportional to current squared
R1
I1 I
V
Rw
~
~ fuse
Rw represents the resistance of the wires
Electrical safety
Short circuit is also a thermal hazard
Short circuit: insulation breaks down Wires touch >> small contact resistance > large current> heating in wires
P =V2/Rs
Rs is extremely small, hence since V is constant Power is very large>> thermal damage
Prevention –include circuit breaker or fuse. Interrupts current if it exceeds specific value
Heat produced can be large >>>>>ignition of adjacent material>> fire
Prevention –include circuit breaker or fuse. Interrupts current if it exceeds specific value
~ V
appliance
Insulating case
2-pin plug
Ground
live
neutral
Electrical safety Shock Hazard Possibility of current passing through person to earth
Impact on person depends mainly on the current
Maximum harmless current (at 50Hz) ≈ 5mA
Some appliances use an insulated two wire system: Appliance has insulating enclosure to prevent direct contact with current. Insulation has very high electrical resistance
Risk reduction
Electrical safety Shock Hazard
Old appliances sometimes used a two wire system insulated from a metal enclosure
Safe unless insulation breaks down
~ V
appliance
metal case 2-pin plug
Ground
live
neutral
Insulation
Electrical safety Shock Hazard If insulation breaks down and live wire contacts the metal case
Any person touching the case is in parallel with the full voltage
~ V
appliance
metal case 2-pin plug
Ground
live
neutral
Ip = V /Rp Current passing through person =
Severity of shock depends on resistance to ground Rp
Bare feet on wet ground > low resistance> very severe
Shoes on rubber mat > high resistance> little affect
Rp
220V/5mA = 44kΩ
It essentially connects the case of appliance through the plugs and sockets to a copper rod inserted into earth outside house. If the live wire touches the metal case, current will go directly to the earth.
~ V
appliance
Grounded Metal case
ELCB
3 pin plug Ground
live
neutral
ground
Live - brown Neutral – blue Earth – green/yellow
Electrical safety Three wire system Additional Earth line for safety
Earth Leakage Circuit Breaker
Electrical safety
~ V
appliance
Grounded metal case
ELCB
3 pin plug Ground
live
neutral
Safety device makes use of Faraday’s law
Coil magnetically senses any difference in the current in the live and neutral wires and activates circuit breaker >5mA
Current in live and neutral wires should be equal unless short circuit occurs in the applicance e.g to Earth
V in
Circuit breaker
Iron ring
Will a 220V hospital circuit protected by a 15A circuit breaker be able to operate a 200W ECG monitor, a 1200W microwave oven and eight 100W lights simultaneously? Veff = 220V Ieff = 15A
The maximum power consumption allowed is: P = IeffVeff = 220V x15A = 3300W
All the apparatus will consume:
P = 200W + 1200W + 8*100W = 2200W
So all apparatus will be able to operate simultaneously.
Example