Lecture 3.1 ELECTRCTY Alternating current Electrical safety
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) Direct voltage (current) dc oltage (& current) are constant t Alternating Current (ac): Nearly all the electricity we use is in the form of alternating current (& voltage) termed ac + t oltage periodic: Magnitude and direction changes - - generally sinusoidal. current also alternates
Alternating Current (ac) + = sinωt t - Angular frequency ω = π f = π / - T + t - = sinωt Average values of the current and voltage is zero Time averaged value of sine function over one or more cycles is zero
+ Alternating Current (ac) = sinωt t - Electrical power consumed by any resistive component in a circuit is P = R dc R ac R + - = sinωt = sinωt 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 > )
Alternating Current (ac) 1 = sinωt = t sin ω t Average value of 1 ( ) = ( ) ave always positive Square root of the average value of the square of the current Root mean square (rms) value 1 rms = = = = Similarly ( ) rms ( ) 1 ave = 1 rms rms = eff eff
Alternating Current (ac) + T = sinωt eff t - is the maximum current value or current amplitude Average current value is zero. T is the period (= 1/f) ac voltage and current are always characterized by their effective (or root mean square ) values: eff = o and eff = o where o, and o are the amplitudes or maximum values of the voltage and the current.
+ Alternating Current T eff t - eff = o and 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/ )amps ac mains varies at a rate of 5 times per second frequency of 5 cycles per second S unit of frequency named after German Physicist Heinrich Hertz 1857-1894. ac mains---frequency 5 Hertz ----5Hz So periodic time (time for one cycle) T= (1/5)s = x1-3 s = millisecs eff = o
+ Alternating Current T eff t - eff = o and eff = o Meters that measure ac current and voltage Calibrated to measure: effective (or rms ) current and voltage eff and eff
Power in an ac circuit Power in a dc circuit ac power P = P= R P = R P ave = eff eff P ave = ( / ) ( / ) P ave = ( o o ) since = R P ave = ( o ) R since = /R P ( o ) ave = R
Alternating Current ESB provides electricity voltage of frequency of 5Hz. The period of the oscillations is: T = 1/5Hz =.s = milliseconds. is the effective or rms value of the voltage: eff = amplitude is: = eff * = 311 Mains voltage swings from +311 to -311 5 times every second giving an effective voltage of. The effective current going through a 1W light bulb is: eff = P ave / eff = (1/) A =.45A
Alternating Current Nearly all the electricity we use is in the form of alternating current (& voltage ) termed ac. Why? 1. ac voltage (compared with dc voltage) can be increased or decreased easily (with a transformer). High voltage can be transmitted more efficiently Transmission of electrical energy from generating stations to homes and businesses. Transmission lines (cables) 1 s km long Therefore resistance of cables is significant Transmission more efficient at high voltages P = R ave eff energy loss is reduced by reducing current
Alternating Current Power loss in transmission cable of resistance R : P loss = R eff Want to deliver a power P to the substation where the transmission line arrives P = = P R R loss eff eff To minimise loss need to increase voltage for a fixed power P to be delivered. Typically 1-7 k Step down transformer used to convert the voltage to for use in home or workplace.
Example A maximum ac voltage of 4 at a frequency 5 Hz is connected across a 53 Ω resistor. Find (a) rms voltage. (b) rms current. (c) the average power and (d) the maximum power dissipated in the resistor. R (a) = rms rms 4 = = 17 (b) rms 17 rms = =.3A R 53Ω (c) rms ( ) = sinωt = sinωt P = R =.3A 53Ω= 54.3Watts ave (d) P max max 17 rms = = = = 19Watts R R 53 f the frequency is doubled, does the average power dissipated in the resistor increase, decrease or stay the same.
Electrical Power Example A 1W bulb operates at a voltage of. Determine the resistance of its filament. P ave = /R or R = /P ave R = () /1 = 484Ω Determine the rms and peak currents through the bulb P ave = rms rms rms Pave 1 = = = rms.455amps ( ) ( ) rms = =.455amps =.643amps
Example Bird sitting on high voltage transmission cable. Why is it not electrocuted? 5 k high voltage transmission cable 1km long, resistance 8 Ω supplying 1 MW of power. Bird s feet are 7 cm apart. Current in the cable P 1MWatts = 4amps = 5k = R Resistance of 7 cm length 7 1 1 1 m m 4 = 8Ω= 19.6 1 Ω 3 oltage between feet 4 = R = 4 19.6 1 =.784 P Assume resistance through body = 8 kω Current through bird = Current through bird = =.784 6 = 98 1 amps 8kΩ 5k = 31.5amps 8kΩ
~ ~ Electrical safety Hazards: Thermal & Electrical shock Thermal Heat produced faster than it can be dissipated Normally thermal energy produced in wires is negligible unless current becomes very large e.g. As more resistances (appliances) are added in parallel current increases R w represents the resistance of the wires R w 1 3 4 5 6 fuse R 1 R R 3 Power dissipated in the feed cable P = R w Power proportional to current squared
Electrical safety Heat produced can be large >>>>>ignition of adjacent material>> fire Prevention include circuit breaker or fuse. nterrupts current if it exceeds specific value Short circuit is also a thermal hazard Short circuit: insulation breaks down Wires touch >> small contact resistance > large current> heating in wires P = /R s R s is extremely small, hence since is constant Power is very large>> thermal damage Prevention include circuit breaker or fuse. nterrupts current if it exceeds specific value
Electrical safety Shock Hazard Possibility of current passing through person to earth mpact on person depends mainly on the current Maximum harmless current (at 5Hz) 5mA Risk reduction Some appliances use an insulated two wire system: Appliance has insulating enclosure to prevent direct contact with current. nsulation has very high electrical resistance appliance live ~ Ground neutral -pin plug nsulating case
Shock Hazard Electrical safety Old appliances sometimes used a two wire system insulated from a metal enclosure Safe unless insulation breaks down live appliance ~ neutral -pin plug metal case Ground nsulation
Electrical safety Shock Hazard f insulation breaks down and live wire contacts the metal case Any person touching the case is in parallel with the full voltage live appliance ~ R p neutral -pin plug metal case Ground /5mA = 44kΩ Current passing through person = p = /R p Severity of shock depends on resistance to ground R p Bare feet on wet ground > low resistance> very severe Shoes on rubber mat > high resistance> little affect
Electrical safety Three wire system Additional Earth line for safety t essentially connects the case of appliance through the plugs and sockets to a copper rod inserted into earth outside house. f the live wire touches the metal case, current will go directly to the earth. appliance live ELCB ~ Ground neutral ground 3 pin plug Grounded Metal case Live - brown Neutral blue Earth green/yellow
Electrical safety Earth Leakage Circuit Breaker appliance live ELCB ~ Ground neutral Safety device makes use of Faraday s law Grounded metal case 3 pin plug in Current in live and neutral wires should be equal unless short circuit occurs in the applicance e.g to Earth ron ring Circuit breaker Coil magnetically senses any difference in the current in the live and neutral wires and activates circuit breaker >5mA
Example Will a hospital circuit protected by a 15A circuit breaker be able to operate a W ECG monitor, a 1W microwave oven and eight 1W lights simultaneously? eff = eff = 15A The maximum power consumption allowed is: P = eff eff = x15a = 33W All the apparatus will consume: P = W + 1W + 8*1W = W So all apparatus will be able to operate simultaneously.