*UNIT 21 : ALTERNATING CURRENT (5 Hours)21.1 Alternating current21.2 Root mean square (rms)21.3 Resistance, reactance and impedance21.4 Power and power factor

**21.1 Alternating Current (1 hour)SUBTOPIC :LEARNING OUTCOMES : Define alternating current (AC).

Sketch and interpret sinusoidal AC waveform.

Write and use sinusoidal voltage and current equations. At the end of this lesson, students should be able to :

*21.1 Alternating current An alternating current (ac) is the electrical current which varies periodically with time in direction and magnitude. An ac circuit and ac generator, provide an alternating current. The usual circuit-diagram symbol for an ac source is .

*

where: Current The output of an ac generator is sinusoidal and varies with time.I : instantaneous current @ current at time t (in Ampere)

*voltageV : instantaneous voltage @ voltage at time t (in Volt) where:

*IoVoT/2T The output of an ac generator is sinusoidal and varies with time.Equation for the current ( I ) :Equation for the voltage ( V ) :

*Terminology in a.c.Frequency ( f )Definition: Number of complete cycle in one second.Unit: Hertz (Hz) or s-1Period ( T )Definition: Time taken for one complete cycle.Unit: second (s)Equation :

Peak (maximum) current ( Io )Definition: Magnitude of the maximum current.Peak (maximum) voltage ( Vo )Definition: Magnitude of the maximum voltage.Angular frequency ( )Equation:

Unit: radian per second (rads-1)

**21.2 Root Mean Square (rms) (1 hour)SUBTOPIC :LEARNING OUTCOMES : Define root mean square (rms), current and voltage for AC source.

Use ,

At the end of this lesson, students should be able to :

21.2 Root mean square (rms)Root mean square current (Irms) is defined as the effective value of a.c. which produces the same power (mean/average power) as the steady d.c. when the current passes through the same resistor.the average or mean value of current in a half-cycle flows of current in a certain direction

* The r.m.s (root mean square) current means the square root of the average value of the current.Root mean square voltage/p.d (Vrms ) is defined as the value of the steady direct voltage which when applied across a resistor, produces the same power as the mean (average) power produced by the alternating voltage across the same resistor.VV=Vo sin t

* The average power, The peak power, Most household electricity is 240 V AC which means that Vrms is 240 V.

*Example 21.2.1A sinusoidal, 60.0 Hz, ac voltage is read to be 120 V by an ordinary voltmeter. What is the maximum value the voltage takes on during a cycle?b) What is the equation for the voltage ?a)b)

*Example 21.2.2A voltage V= 60 sin 120t is applied across a 20 resistor. What will an ac ammeter in series with the resistor read ?b) Calculate the peak current and mean power.a)b)

*Example 21.2.3The alternating potential difference shown above is connected across a resistor of 10 k. Calculatea. the r.m.s. current,b. the frequency,c. the mean power dissipated in the resistor.

*Solution 21.2.3R =10 x 10-3 ,V0 = 200 V and T = 0.04 s anda)b)c)

*Exercise 21.2An ac current is given as I = 5 sin (200t) where the clockwise direction of the current is positive. FindThe peak currentThe current when t = 1/100 s The frequency and period of the oscillation.5 A , 4.55 A, 31.88 Hz, 0.0314 s

**21.3 Resistance, reactance and impedance (2 hours)SUBTOPIC :LEARNING OUTCOMES : Sketch and use phasor diagram and sinusoidal waveform to show the phase relationship between current and voltage for a single component circuit consisting of

i)Pure resistorii)Pure capacitoriii)Pure inductor

At the end of this lesson, students should be able to :

**b) Define and use:i)capacitive reactance,

ii)inductive reactance,

iii)impedance, , and

phase angle,

c) Use phasor diagram to analyse voltage, current, and impedance of series circuit of:i)RCii)RLiii)RLC

*21.3 Resistance, reactance and impedance Phasor is defined as a vector that rotate anticlockwise about its axis with constant angular velocity. A diagram containing phasor is called phasor diagram. It is used to represent a sinusoidal alternating quantity such as current and voltage. It also being used to determine the phase difference between current and voltage in ac circuit.Phasor diagram

*Phasor diagramNOPAo The projection of OP on the vertical axis (Oy) is ON, represents the instantaneous value. Ao is the peak value of the quantity.yy

*Resistance, reactance and impedance

Key Term/MeaningResistance,R Opposition to current flow in purely resistive circuit.Reactance,X Opposition to current flow resulting from inductance or capacitance in ac circuit.Capacitive reactance,Xc Opposition of a capacitor to ac. Inductive reactance,XL Opposition of an inductor to ac. Impedance, ZTotal opposition to ac.(Resistance and reactance combine to form impedance)

*i) Pure Resistor in the AC CircuitPhasor diagramVR

*i) Pure Resistor in the AC Circuit The current flows in the resistor is The voltage across the resistor VR at any instant is

The phase difference between V and I is In pure resistor, the voltage V is in phase with the current I and constant with time.(the current and the voltage reach their maximum values at the same time).

*i) Pure Resistor in the AC Circuit The resistance in a pure resistor is The instantaneous power, The average power,A resistor in ac circuit dissipates energy in the form of heat

*ii) Pure Capacitor in the AC Circuit Pure capacitor means that no resistance and self-inductance effect in the a.c. circuit.Phasor diagramVR

*ii) Pure Capacitor in the AC Circuit When an alternating voltage is applied across a capacitor, the voltage reaches its maximum value one quarter of a cycle after the current reaches its maximum value,( ) The voltage across the capacitor VC at any instant is equal to the supply voltage V and is given by

The charge accumulates on the plates of the capacitor is The current flows in the ac circuit is

*ii) Pure Capacitor in the AC Circuitandor The phase difference between V and I is

*ii) Pure Capacitor in the AC Circuit In pure capacitor, the voltage V lags behind the current I by /2 radians or the current I leads the voltage V by /2 radians. The capacitive reactance in a pure capacitor is The capacitive reactance is defined as

*ii) Pure Capacitor in the AC Circuit The instantaneous power, The average power, For the first half of the cycle where the power is negative, the power is returned to the circuit. For the second half cycle where the power is positive, the capacitor is saving the power.

*Example 21.3.1An 8.00 F capacitor is connected to the terminals of an AC generator with an rms voltage of 150 V and a frequency of 60.0 Hz. Find the capacitive reactance rms current and the peak current in the circuit.ii) Pure Capacitor in the AC CircuitCapacitive reactance, Rms current,Peak current ?

*iii) Pure Inductor in the AC Circuit Phasor diagramVL Pure inductor means that no resistance and capacitance effect in the a.c. circuit.

*iii) Pure Inductor in the AC Circuit When a sinusoidal voltage is applied across a inductor, the voltage reaches its maximum value one quarter of a cycle before the current reaches its maximum value,( ) The current flows in the ac circuit is When the current flows in the inductor, the back emf caused by the self induction is produced and given by

* At each instant the supply voltage V must be equal to the back e.m.f B (voltage across the inductor) but the back e.m.f always oppose the supply voltage V.iii) Pure Inductor in the AC Circuit Hence, the magnitude of V and B ,orwhere0

* The phase difference between V and I isiii) Pure Inductor in the AC Circuit In pure inductor, the voltage V leads the current I by /2 radians or the current I lags behind the voltage V by /2 radians. The inductive reactance in a pure inductor is

*iii) Pure Inductor in the AC Circuit The inductive reactance is defined as The instantaneous power, The average power, For the first half of the cycle where the power is positive, the inductor is saving the power. For the second half cycle where the power is negative, the power is returned to the circuit.

*Example 21.3.2iii) Pure Inductor in the AC CircuitA coil having an inductance of 0.5 H is connected to a 120 V, 60 Hz power source. If the resistance of the coil is neglected, what is the effective current through the coil.Example 21.3.3A 240 V supply with a frequency of 50 Hz causes a current of 3.0 A to flow through an pure inductor. Calculate the inductance of the inductor.

*i) RC in series circuitIn the circuit diagram : VR and VC represent the instantaneous voltage across the resistor and the capacitor.In the phasor diagram : VR and VC represent the peak voltage across the resistor and the capacitor.

*i) RC in series circuitNote

*i) RC in series circuit The total p.d (supply voltage), V across R and C is equal to the vector sum of VR and VC as shown in the phasor diagram.and

*i) RC in series circuit The impedance in RC circuit, From the phasor diagrams,orI leads V by Impedance diagram

*i) RC in series circuitGraph of Z against fZf0R

*i) RC in series circuitExample 21.3.4An alternating current of angular frequency of 1.0 x 104 rad s-1 flows through a 10 k resistor and a 0.10 F capacitor which are connected in series. Calculate the rms voltage across the capacitor if the rms voltage across the resistor is 20 V.Fromand

*ii) RL in series circuit The voltage across the resistor VR and the capacitor VL are

*ii) RL in series circuit The total p.d (supply voltage), V across R and L is equal to the vector sum of VR and VL as shown in the phasor diagram.and

*ii) RL in series circuitImpedance diagram The impedance in RC circuit, From the phasor diagrams,orV leads I by

*Graph of Z against fZf0Rii) RL in series circuit

*iii) RLC in series circuit

*iii) RLC in series circuit

*iii) RLC in series circuit The voltage across the inductor VL , resistor VR and capacitor VC are

*iii) RLC in series circuit The total p.d (supply voltage), V across L, R and C is equal to the vector sum of VL ,VR and VC as shown in the phasor diagram.

*iii) RLC in series circuitImpedance diagram The impedance in RLC circuit, From the phasor diagrams,V leads I by

*Resonance in RLC series circuit Resonance is defined as the phenomenon that occurs when the frequency of the applied voltage is equal to the frequency of the LRC series circuit.Graph of impedance Z, inductive reactance XL, capacitive reactance XC and resistance R with frequency.The series resonancecircuit is used for tuning a radio receiver.

*Resonance in RLC circuitThe graph shows that : at low frequency, impedance Z is large because 1/C is large. at high frequency, impedance Z is high because L is large. at resonance, impedance Z is minimum (Z=R) which is resonant frequencyand I is maximum

*

*A series circuit contains a 50 resistor adjacent to a 200 mH inductor attached to a 0.050F capacitor, all connected across an ac generator with a terminal sinusoidal voltage of 150 V effective.iii) RLC in series circuitWhat is the resonant frequency ? (1.59 kHz)What voltages will be measured by voltmeters across each element at resonance ? (150V,6kV)c) What is the voltage across the series combination of the inductor and capacitor ?Write the equation for the supply voltage at fr.EXERCISE

*Example 21.3.6A 200 resistor, a 0.75 H inductor and a capacitor of capacitance C are connected in series to an alternating source 250 V, fr = 600 Hz. Calculatea. the inductive reactance and capacitive reactance when resonance is occurred.b. the capacitance C.c. the impedance of the circuit at resonance.d. the current flows through the circuit at resonance. e. Sketch the phasor diagram.iii) RLC in series circuit

*Solution 21.3.6R = 200 , L = 0.75 H ,Vrms = 250 V, f = 600 Hz. iii) RLC in series circuita) b) c) Z = R = 200 d) e)

*Exercise 21.3iii) RLC in series circuitA series RLC circuit has a resistance of 25.0 , a capacitance of 50.0 F, and an inductance of 0.300 H. If the circuit is driven by a 120 V, 60 Hz source, calculateThe total impedance of the circuitThe rms current in the circuitThe phase angle between the voltage and the current.64.9 , 1.85 A, 67.3o

**21.4 Power and power factor (1 hour)SUBTOPIC :LEARNING OUTCOMES : Applyi)average power,

ii)instantaneous power,

iii)power factor,

in AC circuit consisting of R, RC, RL and RLC in series

At the end of this lesson, students should be able to :

*21.4 Power and power factor In an ac circuit , the power is only dissipated by a resistance, none is dissipated by inductance or capacitance. Therefore, the real power (Pr) that is used or gone is given by the average power (Pave) i.e :

*Impedance diagram From the diagrams above,.. (2)(2) into (1)V=rms supply voltage(for RLC circuit)

* The term cos is called the power factor. The power factor (cos ) can vary from a maximum of +1 to a minimum of 0. When = 0o (cos =+1) ,the circuit is completely resistive or when the circuit is in resonance (RCL). When = +90o (cos =0) ,the circuit is completely inductive. When = -90o (cos =0) ,the circuit is completely capacitive.

* The power factor can be expressed either as a percentage or a decimal. A typical circuit has a power factor of less than 1 (less than 100%). Example : A motor has a power factor of 80% and the motor consumes 800 W to operate. In order to operate properly, the motor must be supplied with more power than it consumes i.e.1000 W .power factor (80%)800 W (consume)1000 W (supply)

*Example 21.4.1 An oscillator set for 500 Hz puts out a sinusoidal voltage of 100 V effective. A 24.0 resistor, a 10.0F capacitor, and a 50.0 mH inductor in series are wired across the terminals of the oscillator. a) What will an ammeter in the circuit read ? b) What will a voltmeter read across each element ?c) What is the real power dissipated in the circuit?d) Calculate the power supply.e) Find the power factor.f) What is the phase angle?

*Solution 21.4.1 f=500 Hz , V=100 V , R=24.0 , C=10.0F, L=50.0 mH. a)b)c) Real power ?

*Power factor,d)e)f )

*1.A coil having inductance 0.14 H and resistance of 12 is connected to an alternating source 110 V, 25 Hz. Calculatea. the rms current flows in the coil.b. the phase angle between the current and supply voltage.c. the power factor of the circuit.d. the average power loss in the coil.Exercise 21.44.4 A, 61.3o , 0.48, 0.23 kW

*2.A series RCL circuit contains a 5.10 F capacitor and a generator whose voltage is 11.0 V. At a resonant frequency of 1.30 kHz the power dissipated in the circuit is 25.0 W. Calculatea. the inductanceb. the resistance c. the power factor when the generator frequency is 2.31 kHz.Exercise 21.42.94 x 10-3 H , 4.84 , 0.163

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